Question

Calculate the pH of the solution obtained by mixing $200$ml of $0.1$M $C{H}_{3}COOH$ and $50$ml of $0.2$M $Ca(OH{)}_2$. $K_a={10}^{-5}$.

Answer 1

$\underset{0.1-x}{C{H}_{3}COOH+H_{2}O}\rightleftharpoons \underset{x}{C{H}_{2}CO{O}^{-}}+\underset{x}{H^{+}}$

$K_a=\frac{x^2}{0.1-x}$ x is small

$\frac{x^2}{0.1}=k_a\Rightarrow x=\sqrt{0.1}{k}_a$

$\Rightarrow x=\sqrt{0.1}\times {10}^{-5}$

$={10}^{-2}$

$\left[H^{+}\right]={10}^{-2}M$

$Ca(OH{)}_2+H_{2}O\to C{a}^{+2}+2O{H}^{-}$

$\left[O{H}^{-}\right]=0.2\times 2=0.4M$

$H^{+}+O{H}^{-}\to H_{2}O$

milli mole of $\left[H^{+}\right]=200ml\times {10}^{-2}=2mMol$

milli mole of $\left[O{H}^{-}\right]=50\times 0.4=20mmol$

$2mMol$ is comumed

milli Mole $\left[O{H}^{-}\right]=20-2=18mMol$

$\left[O{H}^{-}\right]=\frac{18}{200+50}=\frac{18}{250}=0.072$

$pH=-\log \left[O{H}^{-}\right]+7=7+1.14=8.14$