Question
Ionisation constant of $$C{H_3}COOH$$ is $$1.7 \times {10^{ - 5}}$$ and concentration of $${H^ + }$$ $$ions$$ is $$3.4 \times {10^{ - 4}}.$$ Then, find out initial concentration of $$C{H_3}COOH$$ molecules.
A.
$$3.4 \times {10^{ - 4}}$$
B.
$$3.4 \times {10^{ - 3}}$$
C.
$$6.8 \times {10^{ - 4}}$$
D.
$$6.8 \times {10^{ - 3}}$$
Answer :
$$6.8 \times {10^{ - 3}}$$
Solution :
$$\eqalign{
& C{H_3}COOH \rightleftharpoons C{H_3}CO{O^ - } + {H^ + } \cr
& {K_a} = \frac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}} \cr
& {\text{Given that,}} \cr
& \left[ {C{H_3}CO{O^ - }} \right] = \left[ {{H^ + }} \right] = 3.4 \times {10^{ - 4}}M \cr
& \,\,\,\,\,\,\,\,{K_a}\,{\text{for}}\,C{H_3}COOH = 1.7 \times {10^{ - 5}} \cr} $$
$$C{H_3}COOH$$ is weak acid, so in it $$\left[ {C{H_3}COOH} \right]$$ is equal to initial concentration. Hence,
$$\eqalign{
& 1.7 \times {10^{ - 5}} = \frac{{\left( {3.4 \times {{10}^{ - 4}}} \right)\left( {3.4 \times {{10}^{ - 4}}} \right)}}{{\left[ {C{H_3}COOH} \right]}} \cr
& \left[ {C{H_3}COOH} \right] = \frac{{3.4 \times {{10}^{ - 4}} \times 3.4 \times {{10}^{ - 4}}}}{{1.7 \times {{10}^{ - 5}}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6.8 \times {10^{ - 3}}M \cr} $$