\([OH^-] = C\alpha \) \( = 2 \times {10^{ - 2}} \times \frac{{10}}{{100}} = 2 \times {10^{ - 3}}\)
\(pOH = -log [2 \times 10^{-3}] = -log2 + (- log 10^{-3})\)
\(= -0.3010 + 3 = 2.699 = 2.7\)
\(pH + pOH = 14 \)
\(pH = 14 - 2.7 = 11.3\)
($BaCO_3$ માટે $K_{sp}$ =$ 5 .1 \times 10^{-9}$)