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Ionic Equilibrium question

2013 · Shift 1 · Q4
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  5. /2013 · Shift 1 · Q4

Ionic Equilibrium question

2013 · Shift 1 · Q4

JEE AdvancedChemistryIonic EquilibriumMultiple correct+3 / −0.75
The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is 1/100th of that of a strong acid (HX, 1M), at 25oC. The Ka of HA is
  1. A
    1 ×\times× 10-4
  2. B
    1 ×\times× 10-5
  3. C
    1 ×\times× 10-6
  4. D
    1 ×\times× 10-3
View written solutionFree

Correct answer: A

Step-by-Step Solution

  1. Understand the Reaction and Rate Law: The hydrolysis of methyl acetate (an ester) is an acid-catalyzed reaction. The reaction is: CH3COOCH3+H2O→H+CH3COOH+CH3OHCH_3COOCH_3 + H_2O \xrightarrow{H^+} CH_3COOH + CH_3OHCH3​COOCH3​+H2​OH+​CH3​COOH+CH3​OH The rate of this reaction is dependent on the concentration of the ester and the concentration of the H⁺ ion catalyst. The rate law can be written as: Rate=k[CH3COOCH3][H+]\text{Rate} = k [\text{CH}_3\text{COOCH}_3] [\text{H}^+]Rate=k[CH3​COOCH3​][H+] where k is the rate constant. The initial concentration of methyl acetate is given as 1 M for both cases.

  2. Case 1: Hydrolysis with a Strong Acid (HX) A strong acid, HX, dissociates completely in water. HX→H++X−HX \rightarrow H^+ + X^-HX→H++X− Given that the initial concentration of HX is 1 M, the concentration of hydrogen ions will be: [H+]strong=[HX]=1 M[H^+]_{\text{strong}} = [HX] = 1 \text{ M}[H+]strong​=[HX]=1 M The initial rate of hydrolysis with the strong acid is: (Rate)strong=k[CH3COOCH3][H+]strong=k(1)(1)=k(\text{Rate})_{\text{strong}} = k [\text{CH}_3\text{COOCH}_3] [H^+]_{\text{strong}} = k (1) (1) = k(Rate)strong​=k[CH3​COOCH3​][H+]strong​=k(1)(1)=k

  3. Case 2: Hydrolysis with a Weak Acid (HA) A weak acid, HA, undergoes partial dissociation in water according to the equilibrium: HA⇌H++A−HA \rightleftharpoons H^+ + A^-HA⇌H++A− The initial rate of hydrolysis with the weak acid is: (Rate)weak=k[CH3COOCH3][H+]weak=k(1)[H+]weak=k[H+]weak(\text{Rate})_{\text{weak}} = k [\text{CH}_3\text{COOCH}_3] [H^+]_{\text{weak}} = k (1) [H^+]_{\text{weak}} = k [H^+]_{\text{weak}}(Rate)weak​=k[CH3​COOCH3​][H+]weak​=k(1)[H+]weak​=k[H+]weak​

  4. Relate the Rates and Find [H⁺] for the Weak Acid The problem states that the initial rate with the weak acid is 1/100th of the rate with the strong acid: (Rate)weak=1100(Rate)strong(\text{Rate})_{\text{weak}} = \frac{1}{100} (\text{Rate})_{\text{strong}}(Rate)weak​=1001​(Rate)strong​ Substituting the expressions for the rates from steps 2 and 3: k[H+]weak=1100(k)k [H^+]_{\text{weak}} = \frac{1}{100} (k)k[H+]weak​=1001​(k) The rate constant k cancels out: [H+]weak=1100 M=10−2 M[H^+]_{\text{weak}} = \frac{1}{100} \text{ M} = 10^{-2} \text{ M}[H+]weak​=1001​ M=10−2 M

  5. Calculate the Acid Dissociation Constant (Kₐ) of HA Now we can determine the KaK_aKa​ for the weak acid HA. We use the equilibrium expression for its dissociation. Let's consider the equilibrium for the 1 M HA solution: HA⇌H++A−HA \rightleftharpoons H^+ + A^-HA⇌H++A− From our calculation in step 4, we know the equilibrium concentration of H⁺ is 10−210^{-2}10−2 M. At equilibrium:

    • [H+]=10−2[H^+] = 10^{-2}[H+]=10−2 M
    • Since H⁺ and A⁻ are formed in a 1:1 ratio, [A−]=10−2[A^-] = 10^{-2}[A−]=10−2 M
    • The equilibrium concentration of HA is its initial concentration minus the amount that dissociated: [HA]=1−[H+]=1−10−2=1−0.01=0.99[HA] = 1 - [H^+] = 1 - 10^{-2} = 1 - 0.01 = 0.99[HA]=1−[H+]=1−10−2=1−0.01=0.99 M

    The acid dissociation constant KaK_aKa​ is given by: Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]}Ka​=[HA][H+][A−]​ Substituting the equilibrium concentrations: Ka=(10−2)(10−2)0.99K_a = \frac{(10^{-2})(10^{-2})}{0.99}Ka​=0.99(10−2)(10−2)​ Ka=10−40.99K_a = \frac{10^{-4}}{0.99}Ka​=0.9910−4​ Ka≈1.01×10−4K_a \approx 1.01 \times 10^{-4}Ka​≈1.01×10−4

  6. Compare with Options The calculated value of KaK_aKa​ is approximately 1×10−41 \times 10^{-4}1×10−4. This matches option A.

    • A: 1×10−41 \times 10^{-4}1×10−4
    • B: 1×10−51 \times 10^{-5}1×10−5
    • C: 1×10−61 \times 10^{-6}1×10−6
    • D: 1×10−31 \times 10^{-3}1×10−3

Thus, the correct option is A.

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