What is [ H^+ ]in mol / L of a solution that is 0.20 M in CH_3 COONa and 0.10 M in CH_3 COOH ? ( K_ a . for CH_3 COOH=1.8 × 10^-5 )

Key Idea $\mathrm{CH}_3 \mathrm{COOH}$CH3​COOH (weak acid) and $\mathrm{CH}_3 \mathrm{COONa}$CH3​COONa (conjugated salt) form acidic buffer and for acidic buffer, $$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { salt }]}{[\text { acid }]}$$pH=pKa​+log[ acid ][ salt ]​ and $$\begin{aligned} {\left[\mathrm{H}^{+}\right] } & =-\operatorname{antilog} \mathrm{pH} \\ \mathrm{pH} & =-\log \mathrm{K}_{\mathrm{a}}+\log \frac{[\text { salt }]}{[\text { acid }]} \\ & \left.\quad \because \mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}\right] \\ & =-\log \left(1.8 \times 10^{-5}\right)+\log \frac{(0.20)}{(0.10)} \\ & =4.74+\log 2 \\ & =4.74+0.3010=5.041 \end{aligned}$$[H+]pH​=−antilogpH=−logKa​+log[ acid ][ salt ]​∵pKa​=−logKa​]=−log(1.8×10−5)+log(0.10)(0.20)​=4.74+log2=4.74+0.3010=5.041​ Now, $$\begin{gathered} {\left[\mathrm{H}^{+}\right]=\operatorname{antilog}(-5.045)} \\ =9.0 \times 10^{-6} \mathrm{~mol} / \mathrm{L} \end{gathered}$$[H+]=antilog(−5.045)=9.0×10−6 mol/L​