# Question #8ed71

Nov 10, 2016

Take molarity of acid rain as $2 \times {M}_{{H}_{2} S {O}_{4}}$ since its basicity is 2.
Applying ${M}_{1} {V}_{1} = {M}_{2} {V}_{2}$
${M}_{{H}_{2} S {O}_{4}}$ = $\frac{{M}_{N a O H} \times {V}_{N a O H}}{V} _ \left({H}_{2} S {O}_{4}\right)$
${M}_{{H}_{2} S {O}_{4}}$ = $\frac{0.0811 \times 1.7}{2 \times 20}$ = $0.00344675$