Question

Calculate the degree of ionization of $0.05M$ acetic acid if its $pK$ value is $4.74$. How is the degree of dissociation, affected when its solution is (a) $0.01M$ and (b) $0.1M$ in hydrochloric acid?

• A. (a) Will decrease
  (b) Will decrease
• B. (a) Will decrease (b) Will increase
  
• C. (a) Will increase (b) Will decrease
  
• D. (a) Will increase (b) Will increase
  

Answer 1

The correct option is A (a) Will decrease

(b) Will decrease

$c=0.05Mp{K}_a=4.74p{K}_a=-log\left(K_a\right)K_a=1.82\times {10}^{-5}=c{\alpha }^2\alpha =\surd \frac{K_a}{c}=1.908\times {10}^{-2}Case1-When0.01MHClistaken$

${CH}_{3}COOH⟷H^{+}+{{CH}_{3}COO}^{-}Initialconc:0.05M00Atequilibrium:0.05-x0.01+xx$

$K_a=\frac{\left[{{CH}_{3}COO}^{-}\right]\left[H^{+}\right]}{\left[{CH}_{3}COOH\right]}=\frac{\left(0.01\right)x}{0.05}$

$x=1.82\times {10}^{-5}\times 0.05/0.01=1.82\times {10}^{-3}\times 0.05M\alpha =\frac{Amountofaciddissociated}{Amountofacidtaken}=1.82\times {10}^{-3}Case2-When0.1MHCltaken{K}_a=\frac{\left[{CH}_3{COO}^{-}\right]\left[H^{+}\right]}{\left[{CH}_{3}COOH\right]}=\frac{0.1x}{0.05}x={1.82X10}^{-4}\times 0.05\alpha ={1.82X10}^{-4}\times 0.05/0.05$

$=1.82\times {10}^{-4}$