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Question #8ed71

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Nov 10, 2016

Take molarity of acid rain as $2 \times M_{H_{2} S O_{4}}$ $2timesM_{H_2SO_4}$ since its basicity is 2.

Explanation:

Applying $M_{1} V_{1} = M_{2} V_{2}$ $M_1V_1 = M_2V_2$

$M_{H_{2} S O_{4}}$ $M_(H_2SO_4)$ = $\frac{M_{N a O H} \times V_{N a O H}}{V_{H_{2} S O_{4}}}$ $[M_(NaOH)timesV_(NaOH)]/V_(H_2SO_4)$

$M_{H_{2} S O_{4}}$ $M_(H_2SO_4)$ = $\frac{0.0811 \times 1.7}{2 \times 20}$ $[0.0811times1.7]/(2times20)$ = $0.00344675$ $0.00344675$

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