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Ionic Equilibrium question

2012 · Shift 0 · Q6
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Ionic Equilibrium question

2012 · Shift 0 · Q6

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, Ka of this acid is :
  1. A
    3 ×\times× 10–1
  2. B
    1 ×\times× 10–3
  3. C
    1 ×\times× 10–5
  4. D
    1 ×\times× 10–7
View written solutionFree

Correct answer: C

  1. Given data
  • Initial concentration of acid HQ: C=0.1 MC = 0.1\,\text{M}C=0.1M
  • pH=3\text{pH} = 3pH=3

So, [H+]=10−3 M[H^+] = 10^{-3}\,\text{M}[H+]=10−3M

  1. Ionization of the weak acid

The acid dissociates as: HQ⇌H++Q−\mathrm{HQ \rightleftharpoons H^+ + Q^-}HQ⇌H++Q−

Let the degree of dissociation produce xxx mol/L of H+H^+H+. Then at equilibrium:

  • [H+]=x=10−3[H^+] = x = 10^{-3}[H+]=x=10−3
  • [Q−]=x=10−3[Q^-] = x = 10^{-3}[Q−]=x=10−3
  • [HQ]=0.1−x=0.1−0.001=0.099[HQ] = 0.1 - x = 0.1 - 0.001 = 0.099[HQ]=0.1−x=0.1−0.001=0.099
  1. Expression for KaK_aKa​

Ka=[H+][Q−][HQ]K_a = \frac{[H^+][Q^-]}{[HQ]}Ka​=[HQ][H+][Q−]​

Substitute the values: Ka=(10−3)(10−3)0.099=10−60.099K_a = \frac{(10^{-3})(10^{-3})}{0.099} = \frac{10^{-6}}{0.099}Ka​=0.099(10−3)(10−3)​=0.09910−6​

Ka≈1.01×10−5K_a \approx 1.01 \times 10^{-5}Ka​≈1.01×10−5

So, Ka≈1×10−5K_a \approx 1 \times 10^{-5}Ka​≈1×10−5

  1. Matching with options
  • A: 3×10−13 \times 10^{-1}3×10−1
  • B: 1×10−31 \times 10^{-3}1×10−3
  • C: 1×10−51 \times 10^{-5}1×10−5
  • D: 1×10−71 \times 10^{-7}1×10−7

Hence the correct option is C.

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