Browse Questions

100 mL of $pH$ =6 solution is diluted to 100 mL by water. $pH$ of the solution will increase by

(a) 9

(b) 1

(c) 0.7

(d) 0.3

Before dilution; $[H^+] = 10^{-6} M$
$\therefore [H^+] from acid = 10^{-6} - 10^{-7} = 0.9 \times 10^{-6} M$
We Know that, $M_1V_1 = M_2V_2$
$\Rightarrow 0.9 \times 10^{-6} \times 100 = 1000 \times M_2$
$\Rightarrow M_2 = 0.9 \times 10^{-7} M$

After dilution; Total $[H^+] = [H^+]$ from acid + $[H^+]$ from water = $0.9 \times 10^{-7} M + 10^{-7} M = 1.9 \times 10^{-7} M$
$pH = - log [H^+] = - log 1.9 \times 10^{-7}$ = 7 - log 1.9 = 7 - 0.3 = 6.7

$\therefore$ Increase in pH = 6.7 - 6 = 0.7