Equilibrium Question 873
Question: If acetic acid is 1.0% ionised in its decinormal solution, then pH of its seminormal solution will be
Options:
A) 3.4
B) 4.8
C) 1.6
D) 2.6
Show Answer
Answer:
Correct Answer: D
Solution:
$ CH_3COOH,CH_3CO{O^{-}}+,{H^{+}} $
$ \begin{matrix} Initial,conce\text{.},\frac{1}{10} & 0 & 0 \\ After,t,time\frac{1}{10}-\frac{1}{10}\times \frac{1.0}{100} & \frac{1}{10}\times \frac{1.0}{100} & \frac{1}{10}\times \frac{1.0}{100} \\ ,=0.1 & ,=0.001 & =0.001 \\ \end{matrix} $
$ K=\frac{0.001\times 0.001}{0.1} $
$ =1\times {10^{-5}} $ (i) In case of seminormal solution $ CH_3COOH,CH_3CO{O^{-}},+{H^{+}} $
$ \begin{matrix} Initial,conc\text{.}\frac{1}{2} & 0 & 0 \\ After,t,time\frac{1}{2}-\frac{1}{2}\times \frac{x}{100} & ,\frac{1}{2}\times \frac{x}{100}, & \frac{1}{2}\times \frac{x}{100} \\ \end{matrix} $
$ =0.5 $ [let solution is $ x% $ ionised] $ K=\frac{{{( \frac{1}{2}\times \frac{x}{100} )}^{2}}}{0.5} $
$ =\frac{x^{2}}{20000} $ (ii) From Eqs (i) and (ii) $ \frac{x^{2}}{20000}=1\times {10^{-5}} $
$ x=0.447 $
$ [{H^{+}}]=\frac{1}{2}\times \frac{x}{100} $
$ =\frac{1}{2}\times \frac{0.447}{100}=0.00224 $
$ pH=-\log ,[{H^{+}}] $
$ =-\log ,[0.00224]=2.65 $
BETA
