Question 43 of 69intermediate🔧 ApplyNumerical2 marks

The ionization constant of HF, HCOOH and HCN at 298K are 6.8×10−4,1.8×10−46.8 \times 10^{-4}, 1.8 \times 10^{-4} and 4.8×10−94.8 \times 10^{-9} respectively. Calculate the ionization constants of the corresponding conjugate base.

Correct Answer

The ionization constants of the conjugate bases are: Kb(F−)=1.47×10−11K_b(F^-) = 1.47 \times 10^{-11}, Kb(HCOO−)=5.56×10−11K_b(HCOO^-) = 5.56 \times 10^{-11}, and Kb(CN−)=2.08×10−6K_b(CN^-) = 2.08 \times 10^{-6}.

Exercise: EXERCISES | Q: 6.43 | (Page 45)
For More Understanding

Explanation

The ionization constant of a conjugate base (KbK_b) is related to the ionization constant of its conjugate acid (KaK_a) and the ionic product of water (KwK_w) by the formula Ka×Kb=KwK_a \times K_b = K_w. At 298K, pKw=14pK_w = 14, which means Kw=1.0×10−14K_w = 1.0 \times 10^{-14}. By rearranging the formula to Kb=Kw/KaK_b = K_w / K_a, we can calculate the KbK_b for each given weak acid.

Solution Steps

  1. Step 1: State the relationship between the ionization constants of conjugate acid-base pairs and water. At 298K, Ka×Kb=KwK_a \times K_b = K_w and Kw=1.0×10−14K_w = 1.0 \times 10^{-14}. Therefore, Kb=KwKaK_b = \frac{K_w}{K_a}.

  2. Step 2: Calculate KbK_b for the conjugate base of HF (F−F^-). Given Ka=6.8×10−4K_a = 6.8 \times 10^{-4}, Kb=1.0×10−146.8×10−4=16.8×10−10=1.47×10−11K_b = \frac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}} = \frac{1}{6.8} \times 10^{-10} = 1.47 \times 10^{-11}.

  3. Step 3: Calculate KbK_b for the conjugate base of HCOOH (HCOO−HCOO^-). Given Ka=1.8×10−4K_a = 1.8 \times 10^{-4}, Kb=1.0×10−141.8×10−4=11.8×10−10=5.56×10−11K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}} = \frac{1}{1.8} \times 10^{-10} = 5.56 \times 10^{-11}.

  4. Step 4: Calculate KbK_b for the conjugate base of HCN (CN−CN^-). Given Ka=4.8×10−9K_a = 4.8 \times 10^{-9}, Kb=1.0×10−144.8×10−9=14.8×10−5=2.08×10−6K_b = \frac{1.0 \times 10^{-14}}{4.8 \times 10^{-9}} = \frac{1}{4.8} \times 10^{-5} = 2.08 \times 10^{-6}.

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