\(p H=2 ; H^{+}=10^{-2}=0.01\, M\)
\(\therefore M_{1}=0.1 ; V_{1}=1\)
\(M_{2}=0.01; V_{2}=?\)
From
\(M_{1} V_{1}=M_{2} V_{2} ; 0.1 \times 1=0.01 \times V_{2}\)
\(V_{2}=10\) litre
volume of water added \(=10-1=9\) litre.
$NH_3$ + $H_2O$ $\rightleftharpoons$ $NH_4^ + + O{H^ - }$