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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 43 of 73
Marks: +1, -0
The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 × 10−410^{-4}, 1.8 × 10−410^{-4} and 4.8 × 10−910^{-9} respectively. Calculate the ionization constants of the corresponding conjugate base.
Solution:  
KbK_b = KwKa\frac{K_w}{K_a}
For F−F^{-} , KbK_b = 10−146.8×10−4\frac{10^{-14}}{6.8 \times 10^{-4}} = 1.47 × 10−1110^{-11}
For HCOO−\mathrm{HCOO}^{-} , KbK_b = 10−141.8×10−4\frac{10^{-14}}{1.8 \times 10^{-4}} = 5.6 × 10−1110^{-11}
For CN−\mathrm{CN}^{-} , KbK_b = 10−144.8×10−9\frac{10^{-14}}{4.8 \times 10^{-9}} = 2.08 × 10−610^{-6}
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