Question

The degree of dissociation of acetic acid in the resulting solution and $pH$ of the solution are:

• A. $\alpha =0.0275$%; $pH=1$
• B. $\alpha =0.0275$%; $pH=2$
• C. $\alpha =0.0175$%; $pH=1$
• D. $\alpha =0.0175$%; $pH=2$

Answer 1

The correct option is C $\alpha =0.0175$%; $pH=1$

$Meq.$ of ${CH}_{3}COOH=500\times 0.2=100$
$Meq.$ of $HCl=500\times 0.2=100$
$\therefore$ $\left[HCl\right]=\frac{100}{1000}=0.1$;
$\left[{CH}_{3}COOH\right]=\frac{100}{1000}=0.1$
For ${CH}_{3}COOH$:
$\underset{Beforedissociation0.1Afterdissociation(0.1-X)}{{CH}_{3}COOH}\rightleftharpoons \underset{0X}{{CH}_{3}CO{O}^{-}}+\underset{0.1\left(fromHCl\right)(0.1+X)}{H^{+}}$
$\therefore$ $K_a=\frac{\left[{CH}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[{CH}_{3}COOH\right]}=\frac{X(0.1+X)}{(0.1-X)}$
Due to common ion effect dissociation of ${CH}_{3}COOH$ is very small in presence of $HCl$.
Therefore, $(0.1+X)=0.1$ and $(0.1-X)=0.1$
$\therefore$ $K_a=\frac{X\dot{0}.1}{0.1}$
$\therefore$ $X=K_a=1.75\times {10}^{-5}$
Thus, degree of dissociation
$\alpha =\frac{X}{0.1}=\frac{1.75\times {10}^{-5}}{0.1}$
$=1.75\times {10}^{-4}=0.000175=0.0175$%
Also
$\left[H^{+}\right]=0.1+X=0.1(X<<0.1)$
$\therefore$ $pH=-\log \left[H^{+}\right]=-\log \left[0.1\right]=1$