Question
The degree of dissociation of $$0.1M$$ weak acid $$HA$$ is $$0.5\% .$$ If $$2\,mL$$ of $$1.0\,M\,HA$$ solution is diluted to $$32\,mL$$ the degree of dissociation of acid and $${H_3}{O^ + }\,ion$$ concentration in the resulting solution will be respectively
A.
$$0.02\,{\text{and}}\,3.125 \times {10^{ - 4}}$$
B.
$$1.25 \times {10^{ - 3}}\,{\text{and}}\,0.02$$
C.
$$0.02\,{\text{and}}\,1.25 \times {10^{ - 3}}$$
D.
$$0.02\,{\text{and}}\,8.0 \times {10^{ - 12}}$$
Answer :
$$0.02\,{\text{and}}\,1.25 \times {10^{ - 3}}$$
Solution :
$$\eqalign{
& {\alpha _1} = 0.005 = \sqrt {{K_a} \times {C_1}} \cr
& {\text{Molarity of diluted solution;}}\,2 \times 1 = 32 \times M, \cr
& M = \frac{1}{{16}}\left( {{C_2}} \right) \cr
& {\alpha _2} = \sqrt {\frac{{{K_a}}}{{{C_2}}}} \cr
& = 0.005\sqrt {16} \cr
& = 0.02 \cr
& {H_3}\mathop O\limits^ + = {C_2}{\alpha _2} \cr
& = \frac{{0.02}}{{16}} \cr
& = 1.25 \times {10^{ - 3}}M \cr} $$