Question

The pH of a 1.00 × 10^-2 M solution of cyanic acid (HOCN) is 2.77 at 25°C. Calculate K_a for HOCN from this result.

Short answer

K_a of HOCN at the given data is 3.472 × 10^-4.

Step-by-step solution

Calculating H+ ion Concentration

pH of the solution is given 2.77 at 25°C.

$\begin{array}{l}{\text{pH=-log[H}}^{\text{+}}\text{]}\\ {\text{[H}}^{\text{+}}{\text{]=10}}^{\text{-pH}}\\ {\text{[H}}^{\text{+}}{\text{]=10}}^{\text{-2.77}}\\ {\text{[H}}^{\text{+}}\text{]=1.698}\times {\text{10}}^{\text{-3}}\end{array}$

Calculation of Ka

$\begin{array}{l}{\text{HOCN}}_{\text{(aq)}}\rightleftarrows {{\text{\hspace{0.17em}H}}^{\text{+}}}_{\text{(aq)}}{{\text{+OCN}}^{\text{-}}}_{\text{(aq)}}\\ 0.0100\\ 0.01-\mathrm{x}1.698\times {10}^{-3}\mathrm{x}\end{array}$

That means, x= [H⁺] = [OCN^-] = 1.698×10^-3

$\begin{array}{c}{\text{HOCN}}_{\text{(aq)}}\rightleftarrows {{\text{\hspace{0.17em}H}}^{\text{+}}}_{\text{(aq)}}{{\text{+OCN}}^{\text{-}}}_{\text{(aq)}}\\ 0.0100\\ 0.01-\text{x\hspace{1em}}1.698\times {10}^{-3}\text{\hspace{1em}x}\\ {\text{K}}_{\text{a}}=\frac{\left[{\text{OCN}}^{\text{-}}\right]\left[{\text{H}}^{\text{+}}\right]}{\text{[}HOCN]}\\ =\frac{{(1.698\times {10}^{-3})}^2}{0.01-0.001698}\\ =\frac{2.883\times {10}^{-6}}{8.302\times {10}^{-3}}\\ {\text{K}}_{\text{a}}=3.472\times {10}^{-4}\\ \end{array}$