\(=-\log \mathrm{K}_{\mathrm{b}}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Base}]}\)
\(=-\log 1.8 \times 10^{-5}+\log \frac{0.20}{0.30}\)
\(=-5-0.25+(-0.176)\)
therefore,
\(4.75-0.176=4.57\)
therefore, \(\mathrm{pH}=14-4.57=9.43\)
\(\mathrm{pOH}=\mathrm{pK} \mathrm{b}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Base}]}\)
\(=-\log \mathrm{K}_{\mathrm{b}}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Base}]}\)
\(=-\log 1.8 \times 10^{-5}+\log \frac{0.20}{0.30}\)
\(=-5-0.25+(-0.176)\)
therefore,
\(4.75-0.176=4.57\)
therefore, \(\mathrm{pH}=14-4.57=9.43\)