Question

Calculate the pH of a 0.437 M aqueous solution of hydrocyanic acid (HCN, Ka = 4.0×10-10) and the equilibrium concentrations of the weak acid and its conjugate base.

Answer 1

Answer:

pH=4.88

$[HCN]_{eq}=0.43699M$

$[CN^-]_{eq}=1.322x10^{-5}M$

Explanation:

Hello,

In this case, the undergoing dissociation reaction is:

$HCN(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+CN^-(aq)$

In such a way, the law of mass action becomes:

$Ka=\frac{[H^+]_{eq}[CN^-]_{eq}}{[HCN]_{eq}}$

Which in terms of the change $x$ due to the reaction extent, goes:

$Ka=\frac{(x)(x)}{0.437-x}=4.0x10^{-10}$

Thus, solving for $x$ by either quadratic equation or solver, the results are:

$x_1=1.322x10^{-5}M\\x_2=-1.322x10^{-5}M$

Clearly, the answer is:

$x_1=1.322x10^{-5}M$

In this manner, since $x$ equals the concentration of hydrogen ions, the pH turns out:

$pH=-log([H^+])=-log(1.322x10^{-5})=4.88$

And the concentration of the HCN and the CN⁻:

$[HCN]_{eq}=0.437M-1.322x10^{-5}M=0.43699M$

$[CN^-]_{eq}=1.322x10^{-5}M$

Best regards.