Question

Calculate pH of mixture of $(400mL,\frac{1}{200}MBa(OH{)}_2)+(400mL,\frac{1}{50}MHCl)+(200\text{mL of water})$
$log2=0.3010$

• A. 2.4
• B. 7
• C. more than 7
• D. 4.5

Answer 1

The correct option is A 2.4

Given, $V_{Ba(OH{)}_2}=V_{HCl}=400mL,\left[Ba\right(OH{)}_2]=\frac{1}{200}M,[HCl]=\frac{1}{50}M,V_{H_{2}O}=200mL$
Number of $H^{+}$ ions from acid $N_1{V}_1=400\times \frac{1}{50}=8$
Number of $O{H}^{-}$ ions from base $N_2{V}_2=400\times \frac{1}{200}\times 2=4$
here, $N_1{V}_1>N_2{V}_2$ so solution will be acidic in nature.
so, $\left[H^{+}\right]=\frac{N_1{V}_1-N_2{V}_2}{\text{Total volume}}$
$\left[H^{+}\right]=\frac{(8-4)}{400+400+200}$
$\left[H^{+}\right]=4\times {10}^{-3}$
so, $pH=-log\left[H^{+}\right]=-log4\times {10}^{-3}$
$pH=3-2log2=2.4$