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Chemical Equilibrium question

2021 · 17 Mar · Shift 1 · Q19
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Chemical Equilibrium question

2021 · 17 Mar · Shift 1 · Q19

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
0.01 moles of a weak acid HA (Ka = 2.0 ×\times× 10 −-− 6) is dissolved in 1.0 L of 0.1 M HCl solution. The degree of dissociation of HA is ‾×\underline{\hspace{2cm}}\times​× 10 −-− 5 (Round off to the Nearest Integer). [Neglect volume change on adding HA. Assume degree of dissociation <<1 ]
Numerical answer
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Correct answer: 2

  1. Write the equilibrium for the weak acid

    HA⇌H++A−\mathrm{HA \rightleftharpoons H^+ + A^-}HA⇌H++A−

    For a weak acid, Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]}Ka​=[HA][H+][A−]​

  2. Initial concentrations

    • Moles of HA added = 0.010.010.01 mol in 1.01.01.0 L C=0.01 MC = 0.01\,\text{M}C=0.01M

    • The solution already contains 0.10.10.1 M HCl, which is a strong acid, so [H+]≈0.1 M[H^+] \approx 0.1\,\text{M}[H+]≈0.1M

  3. Let degree of dissociation of HA be α\alphaα

    Then for HA:

    • Dissociated concentration = Cα=0.01αC\alpha = 0.01\alphaCα=0.01α
    • Undissociated concentration = C(1−α)C(1-\alpha)C(1−α)

    Since α≪1\alpha \ll 1α≪1, [HA]≈C=0.01[HA] \approx C = 0.01[HA]≈C=0.01 and [A−]=0.01α[A^-] = 0.01\alpha[A−]=0.01α

    Also, because of the common ion effect from HCl, [H+]≈0.1[H^+] \approx 0.1[H+]≈0.1

  4. Apply the expression for KaK_aKa​

    Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]}Ka​=[HA][H+][A−]​

    Substituting values: 2.0×10−6=(0.1)(0.01α)0.012.0 \times 10^{-6} = \frac{(0.1)(0.01\alpha)}{0.01}2.0×10−6=0.01(0.1)(0.01α)​

    Simplify: 2.0×10−6=0.1α2.0 \times 10^{-6} = 0.1\alpha2.0×10−6=0.1α

    α=2.0×10−60.1=2.0×10−5\alpha = \frac{2.0 \times 10^{-6}}{0.1} = 2.0 \times 10^{-5}α=0.12.0×10−6​=2.0×10−5

  5. Match with the required form

    Degree of dissociation is α=2×10−5\alpha = 2 \times 10^{-5}α=2×10−5

    So the blank is: 2\boxed{2}2​

  6. Comparison with stored answer

    Stored correct answer = 222

    Our derived answer also = 222.

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