Question 58 of 69intermediate🔧 ApplyNumerical3 marks

The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Correct Answer

Given: Concentration of cyanic acid, C=0.1C = 0.1 M and pH=2.34pH = 2.34.

Calculation of [H+][H^+] and Degree of Ionization (α\alpha): The hydrogen ion concentration is calculated as [H+]=10−pH=10−2.34=4.57×10−3[H^+] = 10^{-pH} = 10^{-2.34} = 4.57 \times 10^{-3} M. The degree of ionization (α\alpha) is given by α=[H+]C=4.57×10−30.1=4.57×10−2\alpha = \frac{[H^+]}{C} = \frac{4.57 \times 10^{-3}}{0.1} = 4.57 \times 10^{-2}.

Calculation of Ionization Constant (KaK_a): For a weak acid, Ka=[H+]2CK_a = \frac{[H^+]^2}{C}. Substituting the values, Ka=(4.57×10−3)20.1=20.88×10−60.1=2.09×10−4K_a = \frac{(4.57 \times 10^{-3})^2}{0.1} = \frac{20.88 \times 10^{-6}}{0.1} = 2.09 \times 10^{-4}. Thus, the ionization constant is 2.09×10−42.09 \times 10^{-4} and the degree of ionization is 4.57×10−24.57 \times 10^{-2}.

Exercise: EXERCISES | Q: 6.60 | (Page 46)
For More Understanding

Explanation

The problem requires calculating the hydrogen ion concentration from the given pH, then using it to find the degree of ionization and the ionization constant of the weak acid.

Solution Steps

  1. Step 1: Calculate [H+][H^+] using the given pH. [H+]=10−pH=10−2.34=4.57×10−3[H^+] = 10^{-pH} = 10^{-2.34} = 4.57 \times 10^{-3} M.

  2. Step 2: Calculate the degree of ionization (α\alpha). α=[H+]C=4.57×10−30.1=4.57×10−2\alpha = \frac{[H^+]}{C} = \frac{4.57 \times 10^{-3}}{0.1} = 4.57 \times 10^{-2}.

  3. Step 3: Calculate the ionization constant (KaK_a). Ka=[H+]2C=(4.57×10−3)20.1=2.09×10−4K_a = \frac{[H^+]^2}{C} = \frac{(4.57 \times 10^{-3})^2}{0.1} = 2.09 \times 10^{-4}.

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