Question

Two solutions, A and B, each of 100 L was made by dissolving 4 g of NaOH and 9.8 g of $H_{2}SO_{4}$ in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is

• A. 5.3
• B. 10.6
• C. 9.4
• D. 7.5

Answer 1

Answer: (B)

$M_{\left(\right. H_{2} S O_{4} \left.\right)}=\frac{9.8}{98 \times 100}=10^{- 3}m$
$M_{N a O H}=\frac{4}{40 \times 100}=10^{- 3}m$
Equivalents of resulting solution $\left[\right.\text{O}\text{H}^{-}\left]\right.=\frac{40 \times 1 0^{- 3} - 10 \times 1 0^{- 3} \times 2}{50}=\frac{2}{5}\times 10^{- 3}$
$\text{pOH}=3.\text{39}, \, \text{pH}=10.6$