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

Answers (1)

As per the information given in the question, C =0.08 mol dm-3 and pH =2.85

We know that the HOCl is a weak acid, therefore its dissociation will be written as-

HOCl+H_{2}O\leftrightharpoons H_{3}O^{+}+OCl^{-}

pH = -log [H+]

Thus, -2.85 = log [H+]

[H+] = antilog (-2.85)

[H+] = 1.41 × 10-3

We are aware that in case of a weak mono basic acid-

H^{+}=\sqrt{K_{a}C}

[H^{+}]^{2}=K_{a}C (squaring both sides)

K_{a}=\frac{[H^{+}]^{2}}{C}=\frac{(1.41\times 10^{-3})^{2}}{0.08}

Thus, K_{a}= 2.5\times 10^{-5}

Therefore, we can conclude that the ionization constant of HOCl will be 2.5\times 10^{-5}

 

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