Solveeit Logo
Login

Question

Question: The ionization constant of HCN is \[4 \times {10^{ - 10}}\]. Calculate the concentration of hydrogen...

The ionization constant of HCN is 4×10−104 \times {10^{ - 10}}. Calculate the concentration of hydrogen ions in a 0.2M solution in HCN containing 1 molL−11\,mol{L^{ - 1}} of KCN.

Explanation

Solution

Hint: To solve this question we must know the relation between equilibrium constant of an acid and the concentration of its ions. To get a clearer idea of the same think of the questions to what degree will a compound produce ions in water? In other words, to what extent will this ionization occur in aqueous medium?

Complete step by step answer:
We must remember that the acid ionization represents the fraction of the original acid that has been ionized in solution. It is actually a reflection of the strength of an acid. Strong acids and bases are those which completely dissociates into ions when placed in an aqueous solution.
We can represent the dissociation of HCN as follows:
HCN→H++CN−HCN \to {H^ + } + C{N^ - }
On applying the law of mass action,
Ka=[H+][CN−][HCN]{K_a} = \dfrac{{\left[ {{H^ + }} \right]\left[ {C{N^ - }} \right]}}{{\left[ {HCN} \right]}}
This can be written as:
[H+]=Ka[HCN][CN−]\left[ {{H^ + }} \right] = \dfrac{{{K_a}\left[ {HCN} \right]}}{{\left[ {C{N^ - }} \right]}}
We should note that in presence of a strong electrolyte, the total CN−C{N^ - }concentration is from KCN which has undergone complete dissociation. We can further assume that the dissociation of HCN is very small and the concentration of HCN can be used as the concentration as undissociated HCN.
Therefore, we may now write:
[HCN]=0.2M\left[ {HCN} \right] = 0.2M and [CN−]=1M\left[ {C{N^ - }} \right] = 1M
If we now substitute these values in the equation we have obtained above,
[H+]=Ka[HCN][CN−]\left[ {{H^ + }} \right] = {K_a}\dfrac{{\left[ {HCN} \right]}}{{\left[ {C{N^ - }} \right]}}
That means
[H+]=(4×10−10)×0.21\left[ {{H^ + }} \right] = \dfrac{{\left( {4 \times {{10}^{ - 10}}} \right) \times 0.2}}{1}
=8×10−11 molL−1= 8 \times {10^{ - 11}}\,mol{L^{ - 1}}
Hence, the correct answer is. 8×10−11 molL−18 \times {10^{ - 11}}\,mol{L^{ - 1}}

Note: When KCN is not present, the hydrogen ion concentration is equal to CK\sqrt {CK} which means 0.2×4×10−10=8.94×10−5molL−1\sqrt {0.2 \times 4 \times {{10}^{ - 10}}} = 8.94 \times {10^{ - 5}}mol{L^{ - 1}}. This shows that the concentration of hydrogen ions fails considerably when KCN is added to HCN solution.