1. Tardigrade
  2. Question
  3. Chemistry
  4. At 25° C, acid dissociation constant of HCN is 4.9× 10- 10M. Calculate the % increase in degree of dissociation on diluting HCN solution from 0.1M to 0.001M. Represent the answer as sum of digits till you get single digit answer.

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Q. At $25^\circ C,$ acid dissociation constant of $HCN$ is $4.9\times 10^{- 10}M.$ Calculate the $\%$ increase in degree of dissociation on diluting $HCN$ solution from $0.1M$ to $0.001M.$ Represent the answer as sum of digits till you get single digit answer.

NTA AbhyasNTA Abhyas 2022

Solution:

$K_{a}=Cα^{2}$
$\alpha _{i}=\sqrt{\frac{4 . 9 \times 10^{- 10}}{0 . 1}}=7\times 10^{- 5}$
$\alpha _{F}=\sqrt{\frac{4 . 9 \times 10^{- 10}}{10^{- 3}}}=\sqrt{49 \times 10^{- 8}}=7\times 10^{- 4}$
$\frac{\alpha _{F} - \alpha _{i}}{\alpha _{i}}\times 100=\frac{7 \times 10^{- 4} - 7 \times 10^{- 5}}{7 \times 10^{- 5}}\times 100=900\%$