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JEE Main 2016 · Chemistry · Electrochemistry · medium

A cell is made by joining a hydrogen electrode Pt | H2(1 bar) | HA(0.10 M) to a standard hydrogen electrode through a salt bridge at 298 K. The…

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A cell is made by joining a hydrogen electrode Pt ∣ H2(1 bar) ∣ HA(0.10 M)\mathrm{Pt\,|\,H_2(1\ bar)\,|\,HA(0.10\ M)} to a standard hydrogen electrode through a salt bridge at 298 K298\ \mathrm{K}. The measured emf is 0.177 V0.177\ \mathrm{V}. Taking 2.303RT/F=0.059 V2.303RT/F=0.059\ \mathrm{V} and treating the acid HA as weak and monobasic, the dissociation constant of HA is nearly
  1. A. 1.0×10−61.0\times10^{-6}
  2. B. 1.0×10−51.0\times10^{-5}
  3. C. 1.0×10−31.0\times10^{-3}
  4. D. 3.0×10−33.0\times10^{-3}
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Answer: (B) 1.0×10−51.0\times10^{-5}

The electrode reaction H++e−→12H2\mathrm{H^+ + e^-\to \tfrac12 H_2} gives E=−0.059 pHE = -0.059\,\mathrm{pH} for the test half-cell; versus SHE the emf is 0.059 pH=0.1770.059\,\mathrm{pH}=0.177, so pH=3\mathrm{pH}=3 and [H+]=10−3[\mathrm{H^+}]=10^{-3} M. For a weak acid, Ka=[H+]2/c=(10−3)2/0.10=1.0×10−5K_a=[\mathrm{H^+}]^2/c=(10^{-3})^2/0.10=1.0\times10^{-5}. Forgetting to divide by the concentration gives 10−610^{-6}; using [H+][\mathrm{H^+}] itself or pH\mathrm{pH} misread as 2.52.5 gives the other options.
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