Question

A volume of 20 ml of 0.8 M $HCN$ solution is mixed with 80 ml of 0.4 M $NaCN$ solution. Calculate the pH of the resulting solution. ${{K}_{a}}$ of $HCN$= $2.5\times {{10}^{-10}}$. (log 2 = 0.3)
A. 9.9
B. 9.3
C. 4.1
D. 4.7

Answer 1

Hint: pH is defined as power of hydrogen concentration. It generally tells us about the acidity or basicity of an aqueous solution. If the pH value is less than 7 then the solution is acidic in nature, pH is greater than 7 then it is said to be basic in nature and if the pH value is exactly 7 then it is neutral in nature.

Complete step by step solution:
${{K}_{a}}$ values are most useful in predicting whether the given species is capable of donating or accepting electron pairs. It generally describes the degree of ionization of an acid and behaves as true indicators of acid strength because adding water to a solution will not change the equilibrium constant.
Given conditions according to question:
A volume of 20 ml of 0.8 M $HCN$ solution is mixed with 80 ml of 0.4 M $NaCN$ solution and
${{K}_{a}}$ of $HCN$= $2.5\times {{10}^{-10}}$
Now we know that pH can be calculated by using the formula
$pH={{\log }_{10}}\dfrac{[Salt]}{[Acid]}-\log {{k}_{a}}$
Now put the values given in question
$pH={{\log }_{10}}\dfrac{[0.4\times 80]}{[0.8\times 20]}-\log (2.5\times {{10}^{-10}})$
$pH={{\log }_{10}}(2)-\log (2.5\times {{10}^{-10}})$
log 2 is given i.e. 0.3 and log $2.5\times {{10}^{-10}}$is -9.6
$\therefore pH=0.3+9.6=9.9$

Hence option A is the correct answer.

Note: Larger the value of ${{K}_{a}}$ stronger will be the acid because it defines that the acid is largely dissociated into its ions and large value of ${{K}_{a}}$ also means the formation of products in the reaction is favored. On the other hand lower value of ${{K}_{a}}$ defines little amount of the acid dissociates i.e. acid given is weak acid.