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Question: What is the \[pH\]of a \[1M\text{ }HCN\]solution,\[Ka={{10}^{-10}}\]?...

What is the pHpHof a 1M HCN1M\text{ }HCNsolution,Ka=10−10Ka={{10}^{-10}}?

Explanation

Solution

Hint : We know that Buffer solution is a mixture of a weak acid and its conjugate base or weak base and its conjugate acid. The pH of the solution changes by little value when a small amount of strong acid or base is added to the buffer solution. The Henderson-Hasselbalch equation relates the pH with dissociation constant and with the concentration of acid and its conjugate base

Complete Step By Step Answer:
The Henderson-Hasselbalch equation establishes the relation between pH or pOH with a dissociation constant of weak acid or base with the concentration of acid and its conjugate base or a weak base and its conjugate acid. The given reaction is as follows: HCN ⇌ H+ + CN−HCN~\rightleftharpoons ~{{H}^{+}}~+~C{{N}^{-}}
pKa values are quantitative measurements for the strength of the acid. A weak acid has pKa values in the range of −2 to 12-2\text{ }to\text{ }12for strong acid it is less than −2-2. In solving such types of questions always remember to add the volumes of acid and its conjugate base and thus it can be given as Ka = [H+] [CN−] [HCN]=10−10Ka~=\text{ }\dfrac{\left[ {{H}^{+}} \right]\text{ }\left[ C{{N}^{-}} \right]~}{\left[ HCN \right]}=10-10
Now, for HCNHCN we have; InitialHCN=1MInitia{{l}_{HCN}}=1M; ΔHCN=−xM{{\Delta }_{HCN}}=-xM and EquilibriumHCN=(1−x)MEquilibriu{{m}_{HCN}}=\left( 1-x \right)M
Similarly for, H+{{H}^{+}} we have InitialH+=0MInitia{{l}_{{{H}^{+}}}}=0M; ΔH+=+xM{{\Delta }_{{{H}^{+}}}}=+xM and EquilibriumH+=xMEquilibriu{{m}_{{{H}^{+}}}}=xM
and for CN−C{{N}^{-}} we have InitialCN−=0MInitia{{l}_{C{{N}^{-}}}}=0M; ΔCN−=+xM{{\Delta }_{C{{N}^{-}}}}=+xM and EquilibriumCN−=xMEquilibriu{{m}_{C{{N}^{-}}}}=xM
The above set is derived from given data and some values are predefined values for associated compound and ions.
Thus, Ka = [x] [x] [1−x]=10−10⇒x21=10−10Ka~=\text{ }\dfrac{\left[ x \right]\text{ }\left[ x \right]~}{\left[ 1-x \right]}={{10}^{-10}}\Rightarrow \dfrac{{{x}^{2}}}{1}={{10}^{-10}}
Here the xx value can be ignored, and therefore xx becomes zero in [1−x];\left[ 1-x \right];
Thus, it becomes x=10−10Mx=\sqrt{{{10}^{-10}}}M
pH=−logx⇒pH=−log10−10⇒pH=5pH=-logx\Rightarrow pH=-log\sqrt{{{10}^{-10}}}\Rightarrow pH=5
Therefore, the pHpHof a 1M HCN1M\text{ }HCNsolution,Ka=10−10Ka={{10}^{-10}} is 5.5.

Note :
Remember that Limitations of Henderson-Hasselbalch equation: It is assumed that at equilibrium the concentration of acid and its conjugate base remains constant. The hydrolysis of water and its dependency on the pH for the solution was neglected. For the sake of simplicity, Hydrolysis of base and acid dissociation was neglected.