Question

The equilibrium constant of a strong acid $\left(HA\right)$ with weak base $BOH$ is ${10}^{11}$. The $pH$ of $0.10M$ solution of $BA$ is

Answer 1

$H^{+}+BOH\rightleftharpoons B^{+}+H_{2}O$
0 0 0.10
x x $0.10-x$

$\frac{\left[B^{+}\right]}{\left[H^{+}\right]\left[BOH\right]}=K$

$\frac{0.10-x}{x^2}={10}^{11}$

$0.10-x\simeq 0.10$

$\frac{0.10}{x^2}={10}^{11}$

$x={10}^{-6}$

$pH=-log\left[H^{+}\right]=-log{10}^{-6}=6$