Question

Calculate pH of solution obtained by mixing $500$ml of $0.4$M, $M_{1}OH$ and $1500$ml of $\frac{0.8}{3}$ molar $N_{2}OH$. $K_{b_1}={10}^{-7}$ $K_{b_2}=4\times {10}^{-7}$

Answer 1

$M_{1}OH\underset{H_{2}O}{\rightleftharpoons }{M}_1++O{H}^{-}$

$0.4-x$ $x$ $x$

$K_{b_1}=\frac{x^2}{0.4-x}$

since x is small

$K_{b_1}=\frac{x^2}{0.4}$

$x^2=0.4K_{b_1}$

$x^2=4\times {10}^{-8}$

$x=2\times {10}^{-4}$

$\left[O{H}^{-}\right]=2\times {10}^{-4}$

$N_{2}OH\underset{H_{2}O}{\rightleftharpoons }{N}_2^{+}+O{H}^{-}$

$\frac{0.8}{3}-y$ $y$ $y$

$K_{b_2}=\frac{y^2}{\frac{0.8}{3}-y}$

$K_{b_2}=\frac{{34}^2}{0.8}$

$\frac{{34}^2}{0.8}=4\times {10}^{-7}$

$4^2=\frac{3.2}{3}\times {10}^{-7}\Rightarrow y=3.26\times {10}^{-4}$

$\left[O{H}^{-}\right]-3.26\times {10}^{-4}$

total amount $O{H}^{-}=2\times {10}^{-4}\times 500ml+3.26\times {10}^{-4}\times 1500ml$

$\left[O{H}^{-}\right]=\frac{2\times {10}^{-4}\times 500+3.26\times {10}^{-4}\times 1500}{2000}$

$={10}^{-4}\times \frac{(1000+4890)}{2000}$

$\left[O{H}^{-}\right]=2.945\times {10}^{-4}$

$pH=7+(-\log (2.945\times {10}^{-4}\left)\right)$

$=7+4-0.469$

$=10.53$