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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 60 of 73
Marks: +1, -0
The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
Solution:  
HCNO ⇌ H++CNO−H^{+} + \mathrm{CNO}^{-}
pH = 2.34 or log [H+][H^{+}] = –2.34 or [H+][H^{+}] = antilog(–2.34) = 4.57 × 10−3M10^{-3} \mathrm{M}
[CNO−][\mathrm{CNO}^{-}] = [H+][H^{+}] = 4.57 × 10−3M10^{-3} \mathrm{M}
KaK_a = (4.57×10−3)(4.57×10−3)0.1\frac{(4.57 \times 10^{-3})(4.57 \times 10^{-3})}{0.1} = 2.09 × 10−410^{-4}
α = KaC\sqrt{\frac{K_a}{C}} = 2.09×10−40.1\sqrt{\frac{2.09 \times 10^{-4}}{0.1}} = 0.0457
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