Question
What is $$\left[ {{H^ + }} \right]$$ in $$mol/L$$ of a solution that is $$0.20\,M$$ in $$C{H_3}COONa$$ and $$0.10\,M$$ in $$C{H_3}COOH?$$
$$\left( {{K_a}\,{\text{for}}\,C{H_3}COOH = 1.8 \times {{10}^{ - 5}}} \right)$$
A.
$$3.5 \times {10^{ - 4}}$$
B.
$$1.1 \times {10^{ - 5}}$$
C.
$$1.8 \times {10^{ - 5}}$$
D.
$$9.0 \times {10^{ - 6}}$$
Answer :
$$9.0 \times {10^{ - 6}}$$
Solution :
Key Idea $$C{H_3}COOH$$ ( weak acid ) and $$C{H_3}COONa$$ ( conjugated salt ) form acidic buffer and for acidic buffer,
$$\eqalign{
& pH = p{K_a} + \log \frac{{\left[ {{\text{salt}}} \right]}}{{\left[ {{\text{acid}}} \right]}} \cr
& {\text{and}}\,\,\left[ {{H^ + }} \right] = - {\text{antilog}}\,\,pH \cr} $$
$$pH = - {\text{log}}\,{K_a} + {\text{log}}\frac{{\left[ {{\text{salt}}} \right]}}{{\left[ {{\text{acid}}} \right]}}$$ $$\left[ {\because \,p{K_a} = - {\text{log}}\,{K_a}} \right]$$
$$\eqalign{
& = - {\text{log}}\left( {1.8 \times {{10}^{ - 5}}} \right) + \log \frac{{\left( {0.20} \right)}}{{\left( {0.10} \right)}} \cr
& = 4.74 + {\text{log}}2 \cr
& = 4.74 + 0.3010 \cr
& = 5.041 \cr
& {\text{Now,}}\,\,\left[ {{H^ + }} \right] = {\text{antilog}}\left( { - 5.045} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 9.0 \times {10^{ - 6}}\,mol/L \cr} $$