Question
What will be the value of $$pH$$ of $$0.01\,mol\,d{m^{ - 3}}$$ $$C{H_3}COOH\left( {{K_a} = 1.74 \times {{10}^{ - 5}}} \right)?$$
A.
3.4
B.
3.6
C.
3.9
D.
3.0
Answer :
3.4
Solution :
$$C{H_3}COOH + {H_2}O \rightleftharpoons $$ $$C{H_3}CO{O^ - } + {H_3}{O^ + }$$
$${K_a} = \frac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H_3}{O^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}}$$
$${K_a} = \frac{{{{\left[ {{H_3}{O^ + }} \right]}^2}}}{{\left[ {C{H_3}COOH} \right]}}$$ $$\left( {\because \left[ {C{H_3}CO{O^ - }} \right] = \left[ {{H_3}{O^ + }} \right]} \right)$$
$$\eqalign{
& {\left[ {{H_3}{O^ + }} \right]^2} = {K_a}\left[ {C{H_3}COOH} \right] \cr
& \left[ {{H_3}{O^ + }} \right] = \sqrt {{K_a}\left[ {C{H_3}COOH} \right]} \cr} $$
$${\text{Given:}}\,{K_a} = 1.74 \times {10^{ - 5}},$$ $$\left[ {C{H_3}COOH} \right] = 0.01\,mol\,d{m^{ - 3}}$$
$$\eqalign{
& \left[ {{H_3}{O^ + }} \right] = \sqrt {1.74 \times {{10}^{ - 5}} \times 0.01} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {1.74 \times {{10}^{ - 7}}} \cr
& \left[ {{H_3}{O^ + }} \right] = 4.17 \times {10^{ - 4}} \cr
& pH = - \log \left[ {{H_3}{O^ + }} \right] \cr
& \,\,\,\,\,\,\,\,\, = - \log \left( {4.17 \times {{10}^{ - 4}}} \right) \cr
& \,\,\,\,\,\,\,\,\, = 3.379 \approx 3.4 \cr} $$