Question 57 of 69intermediate🔧 ApplyNumerical3 marks

The ionization constant of propanoic acid is 1.32×10−51.32 \times 10^{-5}. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

Correct Answer

(i) Degree of ionization in 0.05M solution is 1.62×10−21.62 \times 10^{-2} and pH is 3.09

(ii) Degree of ionization in presence of 0.01M HCl is 1.32×10−31.32 \times 10^{-3}

Exercise: EXERCISES | Q: 6.59 | (Page 46)
For More Understanding

Explanation

The degree of ionization of a weak acid is calculated using the approximation α=Ka/c\alpha = \sqrt{K_a/c}. The pH is found using pH=−log⁡[H+]pH = -\log[H^+] where [H+]=cα[H^+] = c\alpha. In the presence of a strong acid like HCl, the common ion H+H^+ suppresses the ionization of the weak acid, and the new degree of ionization is found using Ka=[H+]α1K_a = \frac{[H^+]\alpha}{1}.

Solution Steps

  1. Part 1: Degree of ionization and pH of 0.05M propanoic acid Given: Ka=1.32×10−5K_a = 1.32 \times 10^{-5} and c=0.05c = 0.05 M Using the formula for weak acids: α=Kac\alpha = \sqrt{\frac{K_a}{c}}

    α=1.32×10−50.05=2.64×10−4=1.62×10−2\alpha = \sqrt{\frac{1.32 \times 10^{-5}}{0.05}} = \sqrt{2.64 \times 10^{-4}} = 1.62 \times 10^{-2}

    Now, [H+]=c×α=0.05×1.62×10−2=8.1×10−4[H^+] = c \times \alpha = 0.05 \times 1.62 \times 10^{-2} = 8.1 \times 10^{-4} M

    pH=−log⁡[H+]=−log⁡(8.1×10−4)=4−log⁡(8.1)=4−0.91=3.09pH = -\log[H^+] = -\log(8.1 \times 10^{-4}) = 4 - \log(8.1) = 4 - 0.91 = 3.09

  2. Part 2: Degree of ionization in presence of 0.01M HCl HCl is a strong acid and completely ionizes, providing [H+]=0.01[H^+] = 0.01 M. For propanoic acid: Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]} Let the new degree of ionization be α′\alpha'. Then [A−]=cα′[A^-] = c\alpha' and [HA]≈c[HA] \approx c.

    1.32×10−5=(0.01)(cα′)c=0.01α′1.32 \times 10^{-5} = \frac{(0.01)(c\alpha')}{c} = 0.01 \alpha' α′=1.32×10−50.01=1.32×10−3\alpha' = \frac{1.32 \times 10^{-5}}{0.01} = 1.32 \times 10^{-3}
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