\(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]\)
\(\mathrm{pH}=-\log [\sqrt{55 \times 10^{-14}}]\)
\(=\frac{1}{2}\left[-\log \left(55 \times 10^{-14}\right)\right]\)\(=\frac{1}{2}[-\log 55+14 \log 10]\)
\(=\frac{1}{2}[-1.74+14]=\frac{1}{2}[12.26]=6.13\)
($BaCO_3$ માટે $K_{sp}$ =$ 5 .1 \times 10^{-9}$)