1.

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

Answer»

Solution :pH of HOCl = 2.85
But, -pH = log `[H^+]`
`THEREFORE -2.85=log [H^+]`
`rArr 315 = log [H^+]`
`rArr [H^+]= 1.413xx10^(-3)`
For weak MONOBASIC ACID `[H^+]= SQRT(K_axxC)`
`K_a=[H^+]^2/C=(1.413xx10^(-3))^2/0.08`
`=24.957xx10^(-5) = 2.4957xx10^(-5)`


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