Question

pH of 0.01 M solution of nacn will be given pKa of HCN is 9.2​

Answer 1

HCN ⇌H+ + CN −

Ka = 0.1(1−α) (0.1α)(0.1α)

It is given that pH = 10.5

so, −log([H+ ]) = 5.2[H+ ] = 6.3×10^−6 = 0.1α

α=6.3×10 −5

Ka = 0.1(1−α)(0.1α)(0.1α)

As α<<1 so 1−α=1

Ka = 0.1α^2 = 3.97×10^−10