Answer:
1. After Equivalence Point
2. After Equivalence Point
3. Before Equivalence Point
4. At the Equivalence Point
5. Before Equivalence Point
6. Before Equivalence Point
Explanation:
First let's write down the balanced equation for this reaction:
[tex]HCL + NaOH -->H_2O + NaCL[/tex]
The above equation proves that we need one mole of NaOH to reach the equivalence point with one mole of HCL.
Thus for 100 ml of the 1 mole solution of HCL, we would need a corresponding 100 ml of 1 mole NaOH solution.
So anything less than 100 ml will be before the equivalence point, and anything larger than 100 ml will be after the equivalence point.
The equivalence point will be reached at exactly 100ml.
The answers reflect the above statements.
PLEASE ANSWER ASAP:
The kinetic theory of heat states that heat is the result of the movement of particles in a system. Is this statement True or False
Answer:
TrueExplanation:
Heat is a kind of energy.
The kinetic theory relates the heat with the movement of the particles: the more the particles move, the larger the kinetic energy of the system. The kinetic theory states that heat is the kinetic energy of the particles, atoms or molecules, in a substance, that is transferred from a substance at higher temperature to other substance at lower temperature.
Based on that principle, the kinetic theory explains the changes of phases of the substances in terms of the motion of the particles: the hotter an object the faster the particles move, the more energetic the particles are, and they occupy more space. Thus, when a solid is heated, the particles move faster and it can pass to liquid or gaseous state.
If there are 160 Calories in 28 g in 1 serving of Flamin' Hot Cheetos, how many gram(s) of Cheetos must you burn to raise the temperature of 20.0 mL of water 1oC?
Answer:
3.5 g
Explanation:
The density of water is 1 g/mL. The mass (m) corresponding to 20.0 mL is 20.0 g.
We can calculate the heat (Q) required to raise the temperature of 20.0 mL of water 1 °C (ΔT).
Q = c × m × ΔT = 1 cal/g.°C × 20.0 g × 1 °C = 20 cal
where,
c is the specific heat capacity of water
There are 160 calories in 28 g of Cheetos. The mass that releases 20 cal is:
20 cal × (28 g/160 cal) = 3.5 g
Final answer:
To raise the temperature of 20.0 mL of water by 1°C, 0.35 grams of Flamin' Hot Cheetos must be burned, utilizing the concept of specific heat capacity and energy content of the Cheetos.
Explanation:
The question asks how many grams of Flamin' Hot Cheetos one must burn to raise the temperature of 20.0 mL (which translates to 20.0 grams, assuming the density of water is 1 g/mL) of water by 1°C. To solve this, we first acknowledge that the specific heat capacity of water is about 1 calorie per gram per Celsius degree (1 cal/g°C). Therefore, to raise 20.0 grams of water by 1°C, it requires 20.0 calories. Given the conversion factor that 1 Calorie (kcal, with a capital 'C', equivalent to food calories) = 1000 calories (small 'c'), we deduce that 20.0 calories = 0.02 Kcal.
From the question's premise, 28 grams of Flamin' Hot Cheetos contain 160 Calories (160 Kcal). To find how many grams are needed to produce 0.02 Kcal, we set up a proportion: 160 Kcal per 28 grams = 0.02 Kcal per x grams. Solving for x, we find that x = (0.02 Kcal * 28 grams) / 160 Kcal = 0.35 grams of Flamin' Hot Cheetos are needed to raise the temperature of 20.0 mL of water by 1°C.
The concentration of Mg2+ in a sample of coastal seawater is 1.09 g/kg seawater. What is the molarity of Mg2+ in this seawater with a density of 1.02 g/mL?
Answer: 0.0458 mol/L
Explanation:
in the attachment
The molarity of seawater for Mg²⁺ 0.044844 M
Convert the concentration of Mg²⁺ from g/kg to g/L. Since the density of seawater is 1.02 g/mL, 1 kg of seawater is approximately equivalent to 1 L:
Thus concentration of Mg²⁺ becomes 1.09 g/kg = 1.09 g/L.
Calculate the number of moles of Mg²⁺:
The molar mass of Mg is 24.305 g/mol.
Moles of Mg²⁺ = 1.09 g / 24.305 g/mol = 0.044844 mol.
Since we have 1 liter of solution, the molarity is simply the number of moles per liter. Therefore,
Molarity of Mg²⁺ = 0.044844 M.
In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow reaction in which the bromide ion is replaced by a water molecule, yielding the pinkish-orange complex ion : Co(NH3)5(H2O)3+
Co(NH3)5Br2+Purple(aq)+H2O(l)? Co(NH3)5(H2O)3+Pinkish?orange(aq)+Br?(aq...
The reaction is first order in Co(NH3)5Br2+, the rate constant at 25 ?C is 6.3�10?6 s?1, and the initial concentration of Co(NH3)5Br2+ is 0.100 M.
A; What is its molarity after a reaction time of 19.0h ?
B; How many hours are required for 69% of the Co(NH3)5Br2+ to react?
A. The molarity after a reaction time of 19.00 hours is 0.0988M.
B. 10.12 hours are required for 69% of the [tex][Co(NH_3)_5]Br^{2+}[/tex].
Given data:
Rate constant (k) is [tex]6.3 * 10^{-6} s^{-1}[/tex].
Initial concentration of [tex][Co(NH_3)_5]Br^{2+}[/tex] is 0.100 M.
The rate law for a first-order reaction is R = k * [A]
Where,
R is the rate of the reaction.
k is the rate constant.
[A] is the concentration of the reactant.
A.
To find the molarity after a reaction time of 19.0 hours,
[tex]ln([A]^t / [A]^o) = -kt\\{[A]^t} = [A]^o * e^{-kt}\\{[A]^t} = 0.100 M * e^{-6.3 * 10^{-6} * (19.0 hours * 3600 s/hour)}\\{[A]^t} = 0.0988 M[/tex]
B.
[tex]ln([A]^t / [A]^0) = -kt\\ln(0.69) = -kt[/tex]
Solve for t:
[tex]t = -ln(0.69) / k[/tex]
Substitute the value of k:
[tex]t = -ln(0.69) / (6.3 * 10^{-6} s^{-1})[/tex]
[tex]t = 36441.49 seconds[/tex]
On converting seconds to hours:
t ≈ 10.12 hours
Therefore,
A. The molarity after a reaction time of 19.00 hours is 0.0988M.
B. The time required to react 69% of reactant is 10.12hours.
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The molarity of Co(NH3)5Br2+ after 19 hours is 0.073 M. Approximately 35.75 hours are required for 69% of the Co(NH3)5Br2+ to react.
Explanation:This question pertains to the concept of chemical kinetics, specifically the first order reaction. The molarity after a certain period can be calculated using the formula for first order kinetics: [A] = [A0]e^-kt, where [A0] is the initial concentration, k is the rate constant, t is time, and [A] is the concentration at time t.
A; After 19.0 hours (or 68400 seconds, converted from hours to seconds), the molarity of Co(NH3)5Br2+ can be calculated as follows: [Co(NH3)5Br2+] = (0.100 M)e^-(6.3×10^-6 s^-1 × 68400 s) = 0.073 M.
B; To find the time when 69% of the Co(NH3)5Br2+ has reacted, you need to calculate the time when 31% of the initial molarity remains (as 100% - 69% = 31%). Hence, using the same equation, you can rearrange to find t = -(1/k)ln([A]/[A0]) = -(1/(6.3×10^-6 s^-1)) ln(0.031/0.100) = Approx. 128,680 seconds, or about 35.75 hours.
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Some COCl2 is placed in a sealed flask and heated to 756 K. When equilibrium is reached, the flask is found to contain COCl2 (7.40×10-4 M), CO (3.76×10-2 M), and Cl2 (1.78×10-2 M). What is the value of the equilibrium constant for this reaction at 756 K?
Answer:
[tex]9.044\times 10^{-3}[/tex] is the value of the equilibrium constant for this reaction at 756 K.
Explanation:
[tex]COCl_2\rightleftharpoons CO+Cl_2[/tex]
Equilibrium concentration of [tex]COCl_2[/tex]
[tex][COCl_2]=7.40\times 10^{-4} M[/tex]
Equilibrium concentration of [tex]CO[/tex]
[tex][CO]=3.76\times 10^{-2} M[/tex]
Equilibrium concentration of [tex]Cl_2[/tex]
[tex][Cl_2]=1.78\times 10^{-4} M[/tex]
The expression of an equilibrium constant can be written as;
[tex]K_c=\frac{[CO][Cl_2]}{[COCl_2]}[/tex]
[tex]=\frac{3.76\times 10^{-2}\times 1.78\times 10^{-4}}{7.40\times 10^{-4}}[/tex]
[tex]K_c=9.044\times 10^{-3}[/tex]
[tex]9.044\times 10^{-3}[/tex] is the value of the equilibrium constant for this reaction at 756 K.
uppose you perform a titration of an unknown weak acid solution. You start with 4.00 mL of the weak acid and find that it takes 12.8 mL of 0.0500 M NaOH to reach the equivalence point. What is the concentration of the unknown weak acid solution?
Explanation:
It is known that at the equivalence point,
Equivalent of acid = equivalent of base ......... (A)
and, Equivalent = [tex]\text{Molarity} \times n_{f} \times V[/tex]
Therefore, equivalent of acid is calculated as follows.
Equivalent of acid = [tex]M \times 1 \times 4[/tex] .......... (1)
Equivalent of base = [tex]0.05 \times 1 \times 12.8[/tex] ............ (2)
Hence, using equation (A) we will calculate the concentration as follows.
[tex]M \times 4 = 0.05 \times 12.8[/tex]
M = 0.16 M
Thus, we can conclude that concentration of the unknown weak acid solution 0.16 M.
Consider the reaction below in a closed flask. At 200 o C, the equilibrium constant (Kp) is 2.40 × 103 . 2 NO (g) N2 (g) + O2 (g) If 36.1 atm of NO (g) is added to the closed flask at 200 o C, what is the approximate partial pressure of O2 at equilibrium?
Explanation:
Since, the given reaction is as follows.
[tex]2NO(g) \rightleftharpoons N_{2}(g) + O_{2}(g)[/tex]
Initial: 36.1 atm 0 0
Change: 2x x x
Equilibrium: (36.1 - 2x) x x
Now, expression for [tex]K_{p}[/tex] of this reaction is as follows.
[tex]K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}[/tex]
As the initial pressure of NO is 36.1 atm. Hence, partial pressure of [tex]O_{2}[/tex] at equilibrium will be calculated as follows.
[tex]K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}[/tex]
[tex]2.40 \times 10^{3} = \frac{x \times x}{(36.1 - 2x)^{2}}[/tex]
x = 18.1 atm
Thus, we can conclude that partial pressure of [tex]O_{2}[/tex] at equilibrium is 18.1 atm.
Answer:
partial pressure O2 = 17.867 atm
Explanation:
Step 1: Data given
Temperature = 200 °C
Kp = 2.40 *10^3
Pressure NO = 36.1 atm
Step 2: The balanced equation
2 NO ⇔ N2 + O2
Step 3: The initial pressure
pNO = 36.1 atm
pN2 = 0 atm
pO2 = 0 atm
Step 4: the pressure at the equilibrium
For 2 moles NO we'll have 1 mol N2 and 1 mol O2
pNO = 36.1 - 2X atm
pN2 = X atm
pO2 = X atm
Step 5: Calculate partial pressures
Kp = pN2 * pO2 / (pNO)²
2.40*10³ = x²/(36.1 - 2x)²
48.99 = x/(36.1-2x)
x = 1768.5 -97.98x
x = 17.867
x = partial pressure O2 = 17.867 atm
Phenolphthalein is an indicator that turns from colorless (acidic form) to magenta (basic form) and has a pKa of 9.40. What is the ratio of the magenta phenolphthalein concentration to the colorless phenolphthalein concentration ([magenta phenolphthalein]/[colorless phenolphthalein]) at a pH of 11?
Answer:
40:1 is the ratio of the magenta phenolphthalein concentration to the colorless phenolphthalein concentration.
Explanation:
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
[tex]pH=pK_a+\log(\frac{[magenta(Php)]}{[Php]})[/tex]
We are given:
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of phenolphthalein = 9.40
[tex][magenta(Php)][/tex] = concentration of magenta phenolphthalein
[tex][Php][/tex] = concentration of colorless phenolphthalein
pH = 11
Putting values in above equation, we get:
[tex]11=9.40+\log(\frac{[magenta(Php)]}{[Php]})[/tex]
[tex]\log(\frac{[magenta(Php)]}{[Php]})=11-9.40=1.6[/tex]
[tex]\frac{[magenta(Php)]}{[Php]}=10^{1.6}=39.81 :1 \approx 40:1[/tex]
40:1 is the ratio of the magenta phenolphthalein concentration to the colorless phenolphthalein concentration.
An element, X, can form a chloride (XCl3) and an iodide (XI3). The chloride can be converted quantitatively into the iodide when heated and exposed to excess iodine. If 0.760 grams of XCl3 are heated with iodine, 1.610 g of XI3 are produced. What is the chemical symbol for this element
Answer:
La is the chemical symbol for this element.
Explanation:
[tex]2XCl_3+3I_2\rightarrow 2XI_3+3Cl_2[/tex]
Let the molar mass of [tex]XCl_3[/tex] be M.
Let the molar mass of [tex]XI_3[/tex] be M'.
Moles of [tex]XCl_3=\frac{0.760 g}{M}=n[/tex]
Moles of [tex]XI_3=\frac{1.610g}{M'}=n'[/tex]
According to reaction , 2 moles of [tex]XCl_3[/tex] gives 2 moles of [tex]XI_3[/tex], then n moles [tex]XCl_3[/tex] will give:
[tex]\frac{1}{1}\times n=n[/tex] moles of [tex]XI_3[/tex]
[tex]n=n'[/tex]
[tex]\frac{0.760 g}{M}=\frac{1.610g}{M'}[/tex]
Atomic mass of iodine = 127 g/mol
Atomic mass of chlorine = 35.5 g/mol
Atomic mass of X = x
[tex]\frac{0.760 g}{x+3\times 35.5 g/mol}=\frac{1.610g}{x+3\times 127 g/mol}[/tex]
Solving for x:
x = 138.9 g/mol
The value of atomic mass of X corresponds to compound named lanthanum.
La is the chemical symbol for this element.
Answer:
The chemical symbol for this element is La
Explanation:
Step 1: Data given
Mass of XCl3 = 0.760 grams
Mass of XI3 = 1.610 grams
Molar mass Cl = 35.45 g/mol
Step 2: The balanced equation
2XCl3 + 3I2 → 2XI3 + 3Cl2
The mol ratio for XCl3: XI3 = 2:2 or 1:1
Step 3: Calculate moles
Moles = mass / molar mass
Moles XCl3 = mass XCl3 / molar mass XCl3
Moles XCl3 = 0.760 grams / (X + 3*35.45 g/mol)
Moles XI3 = 1.610 grams / (X + 3*126.9 g/mol)
0.760 grams / (X + 3*35.45 g/mol) = 1.610 grams / (X + 3*126.9 g/mol)
0.760 / (X + 106.35) = 1.610 / (X + 380.7)
0.760 (X +380.7) = 1.610 ( X + 106.35)
0.760X + 289.3 = 1.610X + 171.2
118.1 = 0.85X
X = 138.9 g/mol
If we look for an element with atomic mass of 138.9 g/mol we find Lanthanum (La)
2LaCl3 + 3I2 → 2LaI3 + 3Cl2
Consider the following metals: Ag, Au, Mg, Ni, and Zn.
Which of these metals could be used as a sacrificial anode in the cathodic protection of an underground steel storage tank?
Steel is mostly iron, so use −0.447 V as the standard reduction potential for steel.
Answer:
Mg and Zn
Explanation:
In cathodic protection, the sacrificial anode corrodes instead of the cathode which it protects. The anode is usually higher than the cathode in the electrochemical series. This also means that the reduction potential of the sacrificial anode is more negative than that of the cathode. Consider the reduction potentials of the metals listed in the question:
Mg=-1.185V
Zn= -0.7618V
Ag= +0.7996V
Au= +1.629V
Ni= -0.251V
The reduction potential of the cathode stated in the question is -0.447V hence only magnesium and zinc can function as sacrificial anode.
The metals that can be used as sacrificial anodes in the cathodic protection of an underground steel storage tank are silver (Ag), magnesium (Mg), nickel (Ni), and zinc (Zn).
Explanation:The metals that can be used as sacrificial anodes in the cathodic protection of an underground steel storage tank are the ones with a lower standard reduction potential (E°) than steel. In this case, we need to compare the reduction potentials of steel to the standard reduction potentials of the given metals.
Silver (Ag) has a standard reduction potential of -0.799 V, which is lower than the E° for steel (-0.447 V), so it can be used as a sacrificial anode.Gold (Au) has a standard reduction potential of +1.40 V, which is higher than the E° for steel, so it cannot be used as a sacrificial anode.Magnesium (Mg) has a standard reduction potential of -2.37 V, which is significantly lower than the E° for steel, so it can be used as a sacrificial anode.Nickel (Ni) has a standard reduction potential of -0.257 V, which is lower than the E° for steel, so it can be used as a sacrificial anode.Zinc (Zn) has a standard reduction potential of -0.763 V, which is lower than the E° for steel, so it can be used as a sacrificial anode.Therefore, the metals that could be used as sacrificial anodes in the cathodic protection of an underground steel storage tank are silver (Ag), magnesium (Mg), nickel (Ni), and zinc (Zn).
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Consider the Diels–Alder reaction of cyclopentadiene with p‑benzoquinone. Diels-Alder reaction between cyclopentadiene and p-benzoquinone Select the correct statement about the stoichiometry of this reaction.
A. Multiple molar equivalents of cyclopentadiene and p‑benzoquinone react to form a long polymer chain.
B. One molar equivalent of p‑benzoquinone can react with only one molar equivalent of cyclopentadiene.
C. Two molar equivalents of p‑benzoquinone can react with one molar equivalent of cyclopentadiene.
D. Two molar equivalents of cyclopentadiene can react with one molar equivalent of p‑benzoquinone.
Answer:
The answer is (D) Two molar equivalents of cyclopentadiene can react with one molar equivalent of p‑benzoquinone.
Explanation:
The reaction between p-benzoquinone and cyclopentadiene is a 1,4-cyclo-addition reaction, will follow the general 4n+2 π electrons rule of reaction between a dienophile and a diene.
Hence, two moles of cyclopentadiene added to p-benzoquinone will result in two possible isomers of cis endo and exo adducts (open and closed rings).
Only reaction conditions can be varied to effect yield increase.
Final answer:
The correct statement regarding the stoichiometry of the Diels-Alder reaction of cyclopentadiene with p-benzoquinone is that one molar equivalent of p-benzoquinone reacts with one molar equivalent of cyclopentadiene, highlighting its [4+2] cycloaddition character and specificity.
Explanation:
Considering the Diels–Alder reaction of cyclopentadiene with p-benzoquinone, the correct statement about the stoichiometry of this reaction is: B. One molar equivalent of p-benzoquinone can react with only one molar equivalent of cyclopentadiene. The Diels-Alder reaction is characterized as a [4+2] cycloaddition, where a conjugated diene (cyclopentadiene) interacts with a dienophile (p-benzoquinone) to form a six-membered cyclic compound. This reaction is highly selective and stereospecific, generally taking place between one mole of diene and one mole of dienophile to produce a distinct cyclohexene derivative without creating long polymer chains or requiring multiple moles of either reactant.
The reaction's selectivity and efficiency contribute to its widespread use in synthetic organic chemistry, particularly in the synthesis of complex molecules and natural products. The process does not favor the formation of polymers or the interaction of uneven molar equivalents of reactants, making option B the correct choice for its stoichiometry.
The enthalpy of reaction changes somewhat with temperature. Suppose we wish to calculate ΔH for a reaction at a temperature T that is different from 298 K. To do this, we can replace the direct reaction at T with a three-step process. In the first step, the temperature of the reactants is changed from T to 298 K. ΔH for this step can be calculated from the molar heat capacities of the reactants, which are assumed to be independent of temperature. In the second step, the reaction is conducted at 298 K with an enthalpy change ΔH°. In the third step, the temperature of the products is changed from 298 K to T. The sum of these three enthalpy changes is ΔH for the reaction at temperature T.
An important process contributing to air pollution is the following chemical reaction:
SO2(g) + ½O2(g) → SO3(g)
For SO2(g), the heat capacity cp is 39.9, for O2(g) it is 29.4, and for SO3(g) it is 50.7 J K⁻¹ mol⁻¹.
Calculate ΔH for the preceding reaction at 500 K, using the enthalpies of formation at 298.15 K from Appendix D.
Answer:
-99.8 kJ
Explanation:
We are given the methodology to answer this question, which is basically Kirchhoff law . We just need to find the heats of formation for the reactants and products and perform the calculations.
The standard heat of reaction is
ΔrHº = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants
where ν are the stoichiometric coefficients in the balanced equation, and ΔfHº are the heats of formation at their standard states.
Compound ΔfHº (kJmol⁻¹)
SO₂ -296.8
O₂ 0
SO₃ -395.8
The balanced chemical equation is
SO₂(g) + ½O₂(g) → SO₃(g)
Thus
Δr, 298K Hº( kJmol⁻¹ ) = 1 x (-395.8) - 1 x (-296.8) = -99.0 kJmol⁻¹
Now the heat capacity of reaction will be be given in a similar fashion:
Cp rxn = ∑ ν x Cp of products - ∑ ν x Cp of reactants
where ν is as above the stoichiometric coefficient in the balanced chemical equation.
Cprxn ( JK⁻¹mol⁻¹) = 50.7 - ( 39.9 + 1/2 x 29.4 ) = - 3.90
= -3.90 JK⁻¹mol⁻¹
Finally Δr,500 K Hº = Δr, 298K Hº + CprxnΔT
Δr,500 K Hº = - 99 x 10³ J + (-3.90) JK⁻¹ ( 500 - 298 ) K = -99,787.8
= -99,787.8 J x 1 kJ/1000 J = -99.8 kJ
Notice thie difference is relatively small that is why in some problems it is o.k to assume the change in enthalpy is constant over a temperature range, especially if it is a small range of temperatures.
Consider the reaction: A <=> B. Under standard conditions at equiliubrium, the concentrations of the compounds are [A] = 1.5 M, and [B] = 0.5 M. Keq' for this reaction is ____ and ∆G°' is _____. (You should not need a calculator for this.)
Answer:
See explanation below
Explanation:
In this case, let's write the equation again:
A <------> B Keq = ?
As we are using standard conditions, we can assume we have a temperature of 0 °C (273 K) and 1 atm.
To get the equilibrium constant we only do the following:
Keq = [B] / [A]
However, the problem is asking the reverse equilibrium constant (because of the ' in Keq'), so, we have to do the reverse division:
Keq' = [A]/[B]
Replacing the given values of A and B:
Keq' = 1.5/0.5 = 3
We have the equilibrium constant, we can calculate now the gibbs free energy with the following expression:
ΔG°' = -RTlnKeq'
As Keq' is > 1, the negative logaritm will result into a negative result or a number < 0, so, calculating this we have:
ΔG°' = -8.31 * 273 ln3
ΔG°' = -2.492.34 J
A reaction was performed in which 0.55 g of 2-naphthol was reacted with a slight excess of 1-bromobutane to make 0.32 g of 2-butoxynaphthalene. Calculate the theoretical yield and percent yield for this reaction.
Answer:
Theoretical yield=0.764 g.
Percent yield 42%.
Explanation:
Hello,
In this case the undergoing chemical reaction is shown on the attached picture. In such a way, starting by the given mass of 2-naphtol, one computes the theoretical yielded grams of 2-butoxynaphthalene as shown below, considering their 1 to 1 molar relationship:
[tex]m_{2-butoxynaphthalene}=0.55g2-naphtol*\frac{1mol2-naphtol}{144.17g2-naphtol}*\frac{1mol2-butoxynaphthalene}{1mol2-naphtol}*\frac{200.28g2-butoxynaphthalene}{1mol2-butoxynaphthalene}=0.764g2-butoxynaphthalene[/tex]
Moreover, the percent yield turns out:
[tex]Y=\frac{0.32g}{0.764g}*100\% =42\%[/tex]
Best regards.
Without the balanced chemical equation and molar masses, the theoretical yield of 2-butoxynaphthalene cannot be calculated. Normally, it involves stoichiometric calculations from the starting material. Percent yield is obtained by dividing actual yield by theoretical yield and multiplying by 100%.
Explanation:To calculate the theoretical yield and percent yield for the given reaction, we need to perform a few stoichiometric calculations. However, without the balanced chemical equation and the molar masses, we cannot calculate the theoretical yield for the specific reaction of 2-naphthol and 1-bromobutane to form 2-butoxynaphthalene. If we had the balanced equation, we would calculate the moles of 2-naphthol used, and using stoichiometry, find the moles of 2-butoxynaphthalene that could be formed. Multiplying by the molar mass would give us the theoretical yield in grams. To find percent yield, we would then divide the actual yield (0.32 g) by the theoretical yield and multiply by 100.
Generally speaking, percent yield is the ratio of the actual yield to the theoretical yield multiplied by 100%. It indicates the efficiency of a chemical reaction and can be decreased by incomplete reactions, side reactions, or product recovery losses.
A certain first-order reaction (A→productsA→products) has a rate constant of 9.90×10−3 s−1s−1 at 45 ∘C∘C. How many minutes does it take for the concentration of the reactant, [A][A] , to drop to 6.25%% of the original concentration? Express your answer with the appropriate units
Answer:
4.66667 minutes
Explanation:
Rate constant, k = 9.90×10−3 s−1
Time = ?
Initial concentration, [A]o = 100
Final concentration, [A] = 6.25
The integral rate law for first order reactions is given as;
ln[A] = ln[A]o − kt
kt = ln[A]o - ln[A]
t = ( ln[A]o - ln[A]) / k
t = [ln(100) - ln(6.25)] / 9.90×10−3
t = 2.77 / 9.90×10−3
t = 0.28006 ×103
t = 280 seconds
t = 4.66667 minutes (Upon conversion by dividing by 60)
Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO 3 are produced by the reaction of 6.0 grams of O 2 with 7.0 grams of S. What is the % yield of SO 3 in this experiment
Answer: 95%
Explanation:
We first write down the balanced reaction equation as shown in the image attached. Next we identify the limiting reactant from the given data. Oxygen is the limiting reactant in this case and it is the mass of oxygen reacted that is used in the estimation of the theoretical yield as shown in the image attached. The percentage yield is then calculated according to the formula shown in the image attachment.
6.8. Liquid methyl ethyl ketone (MEK) is introduced into a vessel containing air. The system temperature is increased to 55°C, and the vessel contents reach equilibrium with some MEK remaining in the liquid state. The equilibrium pressure is 1200 mm Hg. (a) Use the Gibbs phase rule to determine how many degrees of freedom exist for the system at equilibrium. State the meaning of your result in your own words. (b) Mixtures of MEK vapor and air that contain between 1.8 mole% MEK and 11.5 mole% MEK can ignite and burn explosively if exposed to a flame or spark. Determine whether or not the given vessel constitutes an explosion hazard.
Answer:
A) Degree of Freedom (DF) = 2 and this means that 2 variables are required to determine the state of the system
B) The vessel doesn't constitute an explosion hazard.
Explanation:
A) Degree of Freedom(DF)=2 + C - π
Where;
π = 2 (liquid and gas phase equilibrium)
C = 2 (air and MEK)
Thus, DF = 2 + 2 - 2 = 2
DF of 2 means that 2 variables are required to determine the state of the system
B) Using Antoine Equation;
Log(Pmek) = 6.97421 - (1209.6/(55 + 216)) = 2.51
Log(Pmek) = 2.51
Thus,Pmek = 10^(2.51) = 323.59mm of Hg or approximately 324mm of Hg.
So, fraction of mek in vapour phase = 324/1200 = 0.27 or 27%
Now this fraction is more than 11.5% in the question.
Therefore, the vessel contains no explosion hazard.
The number of degrees of freedom for the MEK and air system in the vessel at equilibrium is 2, meaning two variables can be varied independently without changing the number of phases. Without knowing the concentration of MEK in the vapor phase, we cannot determine whether the mixture is explosively hazardous.
Explanation:(a) The Gibbs phase rule is given by F = C - P + 2, where F is the number of degrees of freedom, C is the number of components and P is the number of phases. Here, we have two components (MEK and air), and two phases (liquid and vapor) existing in the vessel. So, the number of degrees of freedom F = 2 - 2 + 2 = 2. This means that two independent variables (such as temperature and pressure) can be varied independently without changing the number of phases.
(b) The mixture of MEK vapor and air in the vessel is potentially explosive if it contains between 1.8 mole% MEK and 11.5 mole% MEK. However, without knowing the exact concentration of MEK in the vapor phase, it is impossible to determine whether the mixture is explosively hazardous or not.
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Give the characteristics of a strong acid:
A. ionizes completely in aqueous solution.
B. has equilibrium far to the right.
C. has a polar bond.
D. has a waker bond to hydrogen.
E. all of the above
Answer:
Not D
Explanation:
Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), H2S(aq)⇌HS−(aq)+H+(aq), K1K1K_1 = 9.39×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), HS−(aq)⇌S2−(aq)+H+(aq), K2K2K_2 = 1.45×10−19, what is the equilibrium constant KfinalKfinalK_final for the following reaction? S2−(aq)+2H+(aq)⇌H2S(aq)
Answer: The value of [tex]K_{final}[/tex] for the net reaction is [tex]7.34\times 10^{25}[/tex]
Explanation:
The given chemical equations follows:
Equation 1: [tex]H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^+(aq.);K_1[/tex]
Equation 2: [tex]HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^+(aq.);K_2[/tex]
The net equation follows:
[tex]S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_{final}[/tex]
As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.
The value of equilibrium constant for net reaction is:
[tex]K_{final}=\frac{1}{K_1}\times \frac{1}{K_2}[/tex]
We are given:
[tex]K_1=9.39\times 10^{-8}[/tex]
[tex]K_2=1.45\times 10^{-19}[/tex]
Putting values in above equation, we get:
[tex]K_{final}=\frac{1}{(9.39\times 10^{-8})}\times \frac{1}{(1.45\times 10^{-19})}=7.34\times 10^{25}[/tex]
Hence, the value of [tex]K_{final}[/tex] for the net reaction is [tex]7.34\times 10^{25}[/tex]
The equilibrium constant for the reaction S2−(aq) + 2H+(aq) ⇌ H2S(aq) is 1.361×10−26.
Explanation:To find the equilibrium constant for the reaction S2−(aq) + 2H+(aq) ⇌ H2S(aq), we can use the equilibrium constant expressions for the given reactions and multiply them together. The equilibrium constant for the first reaction (K1) is 9.39×10−8, and for the second reaction (K2) it is 1.45×10−19. So, the equilibrium constant for the overall reaction (Kfinal) is the product of K1 and K2: Kfinal = K1 × K2 = 9.39×10−8 × 1.45×10−19 = 1.361×10−26.
Your standard iron solution is 0.1511 M Fe(II), your dichromate solution is 0.0181 M dichromate, and it took 26.84 mL of your standard iron solution to titrate the excess dichromate in your unknown. What is the chemical oxygen demand of the sample, in units of mg O2/L? (Report your answer with 4 sig figs)
Answer:
[tex]COD=2030\frac{mg}{L}[/tex]
Explanation:
Hello,
In this case, considering the given information the redox reaction is:
[tex]Fe^{+2}+Cr_2O_7^{-2}\rightarrow Fe^{+3}+Cr^{+3}[/tex]
Which properly balanced turns out:
[tex]Fe^{+2}\rightarrow Fe^{+3}+1e^-\\Cr^{+6}_2O_7^{-2}+14H^++6e^-\rightarrow 2Cr^{+3}+7H_2O\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \downarrow\\6Fe^{+2}+Cr_2O_7^{-2}+14H^+\rightarrow 6Fe^{+3}+2Cr^{+3}+7H_2O[/tex]
In such a way, one sees a 6 to 1 molar relationship between the standard iron (II) solution and the dichromate, therefore, by using the following equation it is possible to determine the oxygen as shown below:
[tex]n_{Fe^{+2}}=n_{Cr_2O_7^{-2}}[/tex]
Thus, the moles of iron (II) solution are:
[tex]n_{Fe^{+2}}=0.1511\frac{molFe^{+2}}{L_{sln}}*0.02684L_{sln}=0.00406molFe^{+2}[/tex]
Moreover, the moles of dichormate result:
[tex]n_{Cr_2O_7^{-2}}=0.00406molFe^{+2}*\frac{1molCr_2O_7^{-2}}{6molFe^{+2}}=6.76x10^{-4}}molCr_2O_7^{-2}[/tex]
Thereby, the volume of the sample is:
[tex]V_{sample}=\frac{n_{Cr_2O_7^{-2}}}{M_{Cr_2O_7^{-2}}}=\frac{6.76x10^{-4}molCr_2O_7^{-2}}{0.0181\frac{molCr_2O_7^{-2}}{L_{sln}} } =0.0373L_{sample}[/tex]
Finally, the chemical oxygen demand result:
[tex]COD=6.76x10^{-4}molCr_2O_7^{-2}*\frac{7molO}{1molCr_2O_7^{-2}}*\frac{16gO}{1molO}*\frac{1000mgO}{1gO}*\frac{1}{0.0373L_{sln}} \\COD=2030\frac{mg}{L}[/tex]
Best regards.
Spilling room-temperature water over your skin on a hot day will cool you down. Spilling vegetable oil (of the same temperature as the water) over your skin on a hot day will not. Explain the difference.
Explanation:
Spilling room temperature water over your skin on a hot day will cook the body down. This is because water has a low heat of vapourization( water can evaporate into gas molecules easily). This evaporation causes a cooling effect on the skin surface and it requires heat energy as liquid is converted to gas. This is a typical example of an endothermic process. BUT spilling vegetable oil of the same temperature as the water over the skin will not cool the body down because oil has a high heat of vapourization and doesn't evaporate because it is a viscous liquid and the molecules are tightly bound to each other. I hope this helps.
Consider the reaction C2H4(g) + H2O(g)CH3CH2OH(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.90 moles of C2H4(g) react at standard conditions. S°surroundings = J/K
Answer: The value of [tex]\Delta S^o[/tex] for the surrounding when given amount of ethene gas is reacted is 238.80 J/K
Explanation:
Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.
The equation used to calculate entropy change is of a reaction is:
[tex]\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}][/tex]
For the given chemical reaction:
[tex]C_2H_4(g)+H_2O(g)\rightarrow CH_3CH_2OH(g)[/tex]
The equation for the entropy change of the above reaction is:
[tex]\Delta S^o_{rxn}=[(1\times \Delta S^o_{(CH_3CH_2OH(g))})]-[(1\times \Delta S^o_{(C_2H_4(g))})+(1\times \Delta S^o_{(H_2O(g))})][/tex]
We are given:
[tex]\Delta S^o_{(CH_3CH_2OH(g))}=282.7J/K.mol\\\Delta S^o_{(C_2H_4(g))}=219.56J/K.mol\\\Delta S^o_{(H_2O(g))}=188.82J/K.mol[/tex]
Putting values in above equation, we get:
[tex]\Delta S^o_{rxn}=[(1\times (282.7))]-[(1\times (219.56))+(1\times (188.82))]\\\\\Delta S^o_{rxn}=-125.68J/K[/tex]
Entropy change of the surrounding = - (Entropy change of the system) = -(-125.68) J/K = 125.68 J/K
We are given:
Moles of ethene gas reacted = 1.90 moles
By Stoichiometry of the reaction:
When 1 mole of ethene gas is reacted, the entropy change of the surrounding will be 125.68 J/K
So, when 1.90 moles of ethene gas is reacted, the entropy change of the surrounding will be = [tex]\frac{125.68}{1}\times 1.90=238.80J/K[/tex]
Hence, the value of [tex]\Delta S^o[/tex] for the surrounding when given amount of ethene gas is reacted is 238.80 J/K
You replicate the CEC analysis of a secondary alcohol you performed in the lab using a reverse phase TLC plate producing the TLC plate below. Based on this TLC plate and the mnemonic in your notes, what can you conclude about the stereochemistry of the alcohol? (1 pts)
Answer:
Answer for the question:
You replicate the CEC analysis of a secondary alcohol you performed in the lab using a reverse phase TLC plate producing the TLC plate below. Based on this TLC plate and the mnemonic in your notes, what can you conclude about the stereochemistry of the alcohol? (1 pts)
is given below which explains the best option for the answer.
Explanation:
The enantiomer of the alcohol cannot be determined.
Oxaloacetic acid (2-ketosuccinic acid) is a very important intermediate in metabolism. The compound is involved in the citric acid cycle for energy production within the cell. However, the compound is unstable and slowly decomposes spontaneously. Draw the decomposition products.
Answer:
Carbon dioxide and acetoacetic acid
Explanation:
Oxaloacetic acid is a β-keto acid. It decarboxylates readily via a six-membered cyclic transition state.
The initial products are carbon dioxide and the enol of a ketone.
The enol is unstable and rapidly tautomerizes to the more stable keto form — acetoacetic acid.
Pyruvic acid and carbon dioxide were formed when oxyaloacetic acid decomposes, and the further discussion can be defined as follows:
Oxaloacetic acid: As oxyaloacetic acid first, pyruvic acid and carbon dioxide, are produced.This [tex]\bold{\beta}[/tex]-keto acid is oxaloacetic acid contains. Its six-membered cyclic transition state allows this one to decarboxylate easily.Carbon dioxide as well as the enol of a ketone are the first by-products.This enol is unstable and tautomerizes to the more stable keto form of acetoacetic acid very quickly.Learn more about the Oxaloacetic acid here:
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Consider the reaction 2 SO2(g) + O2(g) <=> 2 SO3(g), which is exothermic as written. What would be the effect on the equilibrium position of removing O2(g)?
Answer:
Removing O₂, means removing one of the reactants and the system would counteract this effect by producing more O₂, thereby shifting the equilibrium position to the left and favouring the backward reaction.
Explanation:
The principle that explains how changes in temperature, Concentration and Pressure of reactants or products of a reaction at equilibrium affect the equilibrium position of the reaction is the Le Chatelier's principle.
The Principle explains that a system/process if a system/process which is at equilibrium is disturbed/perturbed/constrained by one or more changes (in concentration, pressure or temperature), the system would shift the equilibrium position to counteract the effects of this change.
Removing O₂, means removing one of the reactants (changing its concentration) and the system would counteract this effect by producing more O₂, thereby shifting the equilibrium position to the left and favouring the backward reaction.
Final answer:
Removing O₂(g) from the equilibrium of the exothermic reaction 2 SO₂(g) + O₂(g) <=> 2 SO₃(g) shifts the equilibrium towards the reactants' side, increasing the concentration of SO₂(g) and decreasing the concentration of SO₃(g).
Explanation:
The question revolves around the effect of removing O₂(g) from the equilibrium system of the reaction 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), which is exothermic. According to Le Châtelier's Principle, when a change is made to a system in equilibrium, the system adjusts to counteract that change. In this case, removing O₂(g) (oxygen) would decrease its concentration, prompting the equilibrium to shift towards the reactants' side to increase the concentration of O₂(g) and restore equilibrium. This means that the concentration of SO₂(g) will increase, and the concentration of SO₃(g) will decrease as the reaction shifts left to form more reactants.
Given the following proposed mechanism, predict the rate law for the overall reaction. A2 + 2 B → 2 AB (overall reaction) Mechanism A2 ⇌ 2 A fast A + B → AB slow
Answer:
The rate of the over all reaction is ;
[tex]R=K[A_2]^{1/2}[B][/tex]
Explanation:
Step 1 : [tex]A_2\rightleftharpoons 2 A[/tex] fast
Step 2 : [tex]A + B\rightarrow AB[/tex] slow
Equilibrium constant of the reaction in step 1:
[tex]K_1=\frac{[A]^2}{[A_2]}[/tex]....[1]
Overall reaction:
[tex]A_2 + 2 B \rightarrow 2 AB[/tex]
When there is a chemical reaction which taking place in more than 1 step than the rate of the over all reaction is determined by the slowest step occurring during that process;
Here step 2 is slow step, so the rate of the reaction will be;
[tex]R=k[A][B][/tex]..[2]
Putting value of [A] from [1] in [2]:
[tex]R=k\times \sqrt{K_1\times [A_2]}\times [B][/tex]
[tex]K=k\times (K_1)^{1/2}[/tex]
K = rate constant of the reaction
The rate of the over all reaction is ;
[tex]R=K[A_2]^{1/2}[B][/tex]
The overall rate law for the given reaction mechanism is first order in A* (generated from A2) and first order in B, leading to an overall rate law of rate = k[A*][B], with k being the rate constant.
Explanation:The rate law for a chemical reaction can be determined based on the mechanism of the reaction and the rate-determining step. Considering the given mechanism, where A2 dissociates into 2 A rapidly, and then A reacts with B to form AB slowly, the slow step is the rate-determining step. Therefore, the reaction is first order with respect to A* (generated from A2) and first order with respect to B. This is because the slow step involves one molecule of A* and one molecule of B. The overall reaction rate would be expressed as rate = k[A*][B], where k is the rate constant. If we assume the steady state approximation applies, A2 rapidly reaches a steady concentration of A*, and thus its concentration does not directly figure into the rate law for the slow step.
6. How many moles of Cu(NO3)2 are in 25 ml of a 0.35 M solution of Cu(NO3)2? a) How many moles of NO3- are present in 25 ml of the above solution? b) What volume of this solution should be used to get 0.050 moles of Cu(NO3)2?
Explanation:
Below is an attachment containing the solution.
Answer:
0.00875 moles Cu(NO3)2
a) 0.0175 moles NO3-
b) We should use 0.14L
Explanation:
Step 1: Data given
Volume = 25 mL = 0.025 L
Molarity of a Cu(NO3)2 solution = 0.35 M
Step 2: Calculate moles Cu(NO3)2
Moles Cu(NO3)2 = molarity * volume
Moles Cu(NO3)2 = 0.35 M * 0.025 L
Moles Cu(NO3)2 = 0.00875 moles
Step 3: Calculate moles NO3-
Cu(NO3)2 → Cu^2+ + 2NO3-
In 1 mol Cu(NO3)2 we have 2 moles NO3-
For 0.00875 moles Cu(NO3)2 we'll have 2*0.00875 = 0.0175 moles NO3-
Step 4: What volume of this solution should be used to get 0.050 moles of Cu(NO3)2?
Volume = moles / molarity
Volume = 0.050 moles / 0.35 M
Volume = 0.14L
Under appropriate conditions, nitrogen and hydrogen undergo a combination reaction to yield ammonia: N 2 (g) 3H 2 (g) 2NH3 (g) If the reaction yield is 87.5%, how many moles of N 2 are needed to produce 3.00 mol of NH 3
Answer:
We need 1.714 moles N2
Explanation:
Step 1: Data given
The reaction yield = 87.5 %
Number of moles NH3 produced = 3.00 moles
Step 2: The balanced equation
N2(g)+ 3H2(g) →2NH3(g)
Step 3: Calculate moles N2
For 2 moles NH3 produced we need 1 mol N2 and 3 moles H2
This means, if the yield was 100%, for 3.00 moles NH3 produced , we need 1.5 moles N2
For a 87.5 % yield:
we need more N2, increased by a ratio of 100/87.5.
100/87.5 * 1.5 = 1.714 moles N2
A generic Lewis structure is given where Y represents the central atom and the outer atoms are represented by X. The outer atoms are not necessarily all the same element. The central Y atom has one lone pair of electrons. The Y atom is double bonded to an X atom with two lone pairs of electrons. The Y atom is also single bonded to an X atom with three lone pairs of electrons. Which compound is represented by the generic Lewis structure?
The Lewis structure in the question represents the molecule Oxygen difluoride (OF2), where 'Y' is Oxygen with one lone pair of electrons and 'X' is Fluorine, one of which is double bonded to Oxygen while other is single bonded, with two and three lone pairs of electrons respectively.
Explanation:The given generic Lewis structure shows the molecule Oxygen difluoride (OF2), which fulfills the conditions specified. Here, 'Y' can represent an Oxygen atom (O), which has one lone pair of electrons. 'X' can represent Fluorine atoms (F), one of which is double bonded to Oxygen (giving it two lone pairs of electrons), and the other is single bonded to Oxygen (giving it three lone pairs of electrons).
In the Lewis structure for Oxygen difluoride, the bonding and lone pairs of electrons in the molecule are represented through lines and dots respectively. It looks like this:
O = F
|
F
Each Fluorine atom interacts with eight valence electrons: the six in the lone pairs and the two in the single or the double bond. The Oxygen atom also interacts with eight valence electrons according to the Octet Rule, with two from its lone pair, four from its double bond with one Fluorine atom, and two from its single bond with the other Fluorine atom.
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The compound represented by the generic Lewis structure is [tex]\(\text{SO}_2\).[/tex]
To determine the compound, we need to analyze the information given about the Lewis structure:
1. The central atom Y has one lone pair of electrons.
2. The Y atom is double bonded to an X atom, which has two lone pairs of electrons.
3. The Y atom is also single bonded to an X atom, which has three lone pairs of electrons.
Let's start by identifying the possible elements for X and Y based on the valence electrons and the types of bonds and lone pairs described:
- The Y atom has one lone pair and forms a double bond and a single bond with X atoms. This means Y must have at least 4 valence electrons (2 in the lone pair + 1 in each of the two bonds). Y also forms a total of 3 bonds (1 double bond and 1 single bond), which means it can have a maximum of 6 valence electrons (3 bonds x 2 electrons per bond). Therefore, Y must be an element from Group 16 (chalcogens), which typically have 6 valence electrons. The only chalcogen that can form a double bond with an X atom and still have a lone pair is sulfur (S), as oxygen (O) would not have a lone pair if it formed two double bonds, and the heavier chalcogens (Se, Te) are less likely to form double bonds.
- The X atom that is double bonded to Y has two lone pairs, which means it has at least 4 valence electrons (2 in the double bond + 2 in the lone pairs). This X atom cannot have more than 6 valence electrons because it is not forming more than 2 bonds. Therefore, X must be an element from Group 16 (chalcogens) as well. Given that Y is sulfur, the only chalcogen that can form a double bond with sulfur and have two lone pairs is oxygen (O).
The X atom that is single bonded to Y has three lone pairs, which means it has at least 5 valence electrons (1 in the single bond + 2 in the lone pairs). This X atom cannot have more than 7 valence electrons because it is not forming more than 1 bond. Therefore, X must be an element from Group 17 (halogens), which typically have 7 valence electrons. The most common element from this group that forms a single bond with sulfur and has three lone pairs is chlorine (Cl).
However, since we are looking for a compound where both X atoms are the same element, and considering the generic Lewis structure, we can conclude that both X atoms must be oxygen (O), as oxygen can form both single and double bonds with sulfur.
Putting it all together, the central Y atom is sulfur (S), and the outer X atoms are both oxygen (O), leading to the compound[tex]\(\text{SO}_2\),[/tex]which is sulfur dioxide. The Lewis structure of [tex]\(\text{SO}_2\)[/tex] shows sulfur double bonded to one oxygen atom and single bonded to another oxygen atom, with the appropriate number of lone pairs on each atom as described in the question.
You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p K a = 4.20 ) and 0.220 M sodium benzoate. How many milliliters of each solution should be mixed to prepare this buffer?
Answer:
34.5 mL
Explanation:
Given, we ned to prepared buffer solution with pH = 4.00 , volume = 100.0 mL .
[C6H5COOH] = 0.100 M , [C6H5COO-] = 0.120 M , pKa = 4.20
First we need to calculate the ratio of the conjugate base and acid
We know, Henderson Hasselbalch equation
pH = pKa + log [C6H5COO-] /[C6H5COOH]
4.00 = 4.20 + log [C6H5COO-] /[C6H5COOH]
log [C6H5COO-] /[C6H5COOH] = 4.00-4.20
= - 0.20
Antilog from both side
[C6H5COO-] /[C6H5COOH] = 0.631
Now we are given the molarity of each acid and its conjugate base and we need to calculate for volume
Sum of volume = 100 mL = 0.100 L
volume of benzoic acid = x
volume of benzoate = 0.10 -x ,
so Volume of acid + volume of conjugate base = 0.100 L
[C6H5COO-] /[C6H5COOH] = 0.631
[C6H5COO-] = 0.631 * [C6H5COOH]
0.120 (0.1-x) = 0.631 *0.100x
So, x = 0.065
So, volume of benzoic acid = x = 0.0655 L
= 65.5 mL
So, volume of sodium benzoate = 0.1 -x
= 0.1-0.0655
= 0.0345 L
= 34.5 mL
So we need to mix 65.5 mL of 0.100 M benzoic acid and 34.5 mL of 0.120 M sodium benzoate form 100.0 mL of a pH=4.00 buffer solution.
Answer:
Volume benzoic acid= 77.74 mL
Volume sodium benzoate: = 22.26 mL
Explanation:
Step 1: Data given
Volume = 100.0 mL
pH = 4.00
Molarity benzoic acid = 0.100 M
pKa benzoic acid = 4.20
Molarity sodium benzoate = 0.220 M
Step 2
pH = pKa + log([base]/[acid])
4 = 4.2 + log([base]/[acid])
-0.2 = log([base]/[acid])
10^-0.2 = [base]/[acid]
0.63 = [base]/[acid]
Step 3
start with 100 mL 0.1 M benzoic acid
0.100 L* 0.1M = 0.01 moles.
Since base/acid = 0.63, add 0.63*0.01 moles base = 0.0063 moles base.
And 0.0063 moles of a 0.22 M salt solution are x mL:
0.22 moles = 1000 mL
1 mole = 1000/0.22 mL
0.0063 moles = 1000*0.0063/0.22 mL = 28.63 mL .
So a mixture of 100 mL 0.1 M bencoic acid + 28.63 mL of 0.22 benzoate has pH =4.00
Step 4: For a total volume of 100 mL we'll have:
Volume benzoic acid: 100*100/(128.63) = 77.74 mL
Volume sodium benzoate: 28.63*100/(128.63) = 22.26 mL