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NCERT Exemplar · Q32

Q.pH of 0.08 mol dm^-3 HOCl solution is 2.85. Calculate its ionisation constant.

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From the pH we obtain [H+][\text{H}^+], and for the weak acid HOCl, Ka=[H+]2C−[H+]K_a = \dfrac{[\text{H}^+]^2}{C - [\text{H}^+]}. With C=0.08C = 0.08 M and pH =2.85= 2.85, Ka≈2.5×10−5K_a \approx 2.5 \times 10^{-5}.

Hypochlorous acid is a weak monoprotic acid that ionises as:

HOCl⇌H++OCl−\text{HOCl} \rightleftharpoons \text{H}^+ + \text{OCl}^-

Step 1 — find [H+][\text{H}^+] from the pH.

[H+]=10−pH=10−2.85=1.41×10−3 mol dm−3[\text{H}^+] = 10^{-\text{pH}} = 10^{-2.85} = 1.41 \times 10^{-3}\ \text{mol dm}^{-3}

Step 2 — equilibrium concentrations. Each HOCl that ionises gives one H+\text{H}^+ and one OCl−\text{OCl}^-, so [H+]=[OCl−]=1.41×10−3[\text{H}^+] = [\text{OCl}^-] = 1.41 \times 10^{-3} M. The undissociated acid remaining is:

[HOCl]=C−[H+]=0.08−0.00141=0.0786 M[\text{HOCl}] = C - [\text{H}^+] = 0.08 - 0.00141 = 0.0786\ \text{M}

Step 3 — ionisation constant. …

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