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# If ${K_a}$of HCN = $4 \times {10^{ - 10}}$, then the pH of $2.5{\text{ }} \times {\text{ }}{10^{ - 1}}$molar ${\text{HCN(aq)}}$ is:(A) 1(B) 2.5(C) 4(D) 5

Last updated date: 10th Sep 2024
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Hint: Hydrogen cyanide is a weak acid, having pKa of 9.2. Dissociation constant of weak acid tells us that weak acids dissociate partly in a dilute solution. Based on the concentration of the acid, we can determine the dissociation constant.

Formulas used: $[{H^ + }]{\text{ }} = {\text{ }}\sqrt {{K_a}{\text{ }} \times {\text{ }}c}$ and $pH{\text{ }} = {\text{ }} - \log [{H^ + }]{\text{ }}$

Complete step-by-step solution:
Looking at the structure of HCN, carbon has a single bond with hydrogen and triple bond with nitrogen. This makes carbon really stable due to its complete octet. That is why, carbon is reluctant to donate the proton or hydrogen atom and thereby the bond keeps the molecule very stable.
As we know, pH is convenient to express acidity or alkalinity of a solution by referring to the concentration of hydrogen ions only. This can be expressed as
$pH{\text{ }} = {\text{ }} - \log [{H^ + }]{\text{ }}$
For this to be calculated, first we need to find the concentration of hydrogen ion using the given conditions.
Given: dissociation constant ${K_a}$of HCN = $4 \times {10^{ - 10}}$ and concentration, c = $2.5{\text{ }} \times {\text{ }}{10^{ - 1}}$
Therefore, on applying the formula $[{H^ + }]{\text{ }} = {\text{ }}\sqrt {{K_a}{\text{ }} \times {\text{ }}c}$and putting the given values in it, we get
$[{H^ + }]{\text{ }} = {\text{ }}\sqrt {4{\text{ }} \times {\text{ }}{{10}^{ - 10}} \times {\text{ }}2.5{\text{ }} \times {\text{ }}{{10}^{ - 1}}} {\text{ }}$
So, $[{H^ + }]{\text{ }} = {\text{ }}{10^{ - 5}}{\text{ }}$
Now, we will use the pH formula to calculate the pH using the determined hydrogen ion concentration.
$pH{\text{ }} = {\text{ }} - \log [{H^ + }]{\text{ }}$
$pH{\text{ }} = {\text{ }} - \log {\text{ }}{10^{ - 5}}$
$\therefore {\text{ }}pH{\text{ }} = {\text{ }}5$

Hence, the correct answer is (D).

Note: The ${{\text{H}}^{\text{ + }}}$ ion is considered to be the same as ${H_3}{O^ + }$, as it is the hydrogen ion bonded to the water molecule. Due to the positive charge of proton, it gets attracted to the electrons of water molecules. Therefore, we can write, $pH{\text{ }} = {\text{ }} - \log [{H_3}{O^ + }]{\text{ }}$.