Question

100 mL of 0.005 M $H_{2}S{O}_4$ is diluted to 1 L. The resulting pH will be .

• A. 3
• B. 5
• C. 4

Answer 1

The correct option is C 4

M eq. of $H_{2}S{O}_4$ present initially = M eq. of $H_{2}S{O}_4$ after dilution

$N_1{V}_1$ = $N_2{V}_2$
0.005 x 2 x 100 = 1000 x 2 x $N_2$
$N_2$ = ${10}^{-4}$
$\left[H^{+}\right]$ = ${10}^{-4}$
pH= 4