Homework Answers
Solution:
Before addition of NaOH, the pH of the solution contains only HCN.
HCN is a weak acid, it is dissociated in water as:
HCN + H2O = CN- + H3O+
0.421 M --------0.0M--------0.0M (initial)
0.421 -X-----------X-------------X (at equilibrium)
Ka = X .X / 0.421 - X
HCN is a weak acid, therefore, X will be very small in comparison of 0.421 M
Hence, Ka = X . X / 0.421
Ka for HCN = 6.2 x 10^-10
6.2 x 10^-10 = X2 / 0.421M
X2 = 2.61 x 10^-10
X = √2.61 x 10^-10
X= 1.62 x 10^-5 M
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