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JEE Main 2007
Exam
JEE Main
Year
2007
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2007 · Chemistry · Ionic Equilibrium

The first and second dissociation constants of an acid H_2 A are...

The first and second dissociation constants of an acid H2A\mathrm{H}_2 \mathrm{A} are 1.0×10−51.0 \times 10^{-5} and 5.0×10−105.0 \times 10^{-10} respectively. The overall dissociation constant of the acid will be

Options

  1. A5.0×10−55.0 \times 10^{-5}
  2. B5.0×10155.0 \times 10^{15}
  3. C5.0×10−155.0 \times 10^{-15}
  4. D0.0×1050.0 \times 10^5

Correct answer

C. 5.0×10−155.0 \times 10^{-15}

Step-by-step solution

H2A⇌HA−+H+K1=[HA−][H+][H2A]…(1)HA−⇌H++A2−K2=[H+][A2−][HA−]…(2)\begin{array}{ll}\mathrm{H}_2 \mathrm{A} \rightleftharpoons \mathrm{HA}^{-}+\mathrm{H}^{+} & \mathrm{K}_1=\frac{\left[\mathrm{HA}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{H}_2 \mathrm{A}\right]} \quad \dots (1)\\ \mathrm{HA}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{2-} & \mathrm{K}_2=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{2-}\right]}{\left[\mathrm{HA}^{-}\right]}\quad \dots(2)\end{array} For the reaction H2A⇌2H++A2−K=[H+]2[ A2−][H2A]=K1×K2=1×10−5×5×10−10=5×10−15\begin{aligned} & \mathrm{H}_2 \mathrm{A} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{A}^{2-} \\ & \mathrm{K}=\frac{\left[\mathrm{H}^{+}\right]^2\left[\mathrm{~A}^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{A}\right]}=\mathrm{K}_1 \times \mathrm{K}_2 \\ & =1 \times 10^{-5} \times 5 \times 10^{-10} \\ & =5 \times 10^{-15} \end{aligned} Hence, (C) is correct.
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