Question

Equal volumes of 0.1 M potassium hydroxide and 0.1 M sulphuric acid are mixed. The pH of the resulting solution is:
A. 7
B. 0
C. >7
D. <7

Answer 1

Hint: As we know the gram equivalent of acid neutralizes the equal number of gram equivalent of base. So to find the PH of the resulting solution first we have to find out the normality of the resulting solution.

Complete step by step answer:
The formula for normality is given as the product of molarity and basicity/acidity. Suppose the volume of potassium hydroxide is ${V_1}$ and the volume of sulphuric acid is ${V_2}$. Given that the volume of acid and base are equal thus
\[{V_1} = {V_2}\]
In the case of potassium hydroxide, the molarity is given as 0.1M and the acidity of the potassium hydroxide is 1 because it gives one hydroxyl ion. Thus the normality of the potassium hydroxide (${N_1}$) is calculated as:
$0.1\times 1$ = 0.1 N
Whereas in the case of the sulphuric acid, the basicity is 2 as it gives two hydrogen ions. Thus the normality of sulphuric acid (${N_2}$) is calculated as:
$0.1\times 2$ = 0.2 N.
Since the above case normality of sulphuric acid is more than the normality of the potassium hydroxide then the normality of the resulting solution after mixing is given as:
$\dfrac{{{N_1}{V_1} - {N_2}{V_2}}}{{{V_1} + {V_2}}} = {N_3}$
$ \Rightarrow \dfrac{{(0.2 - 0.1) \times V}}{{2V}} = {N_3}$
$ \Rightarrow \dfrac{{0.1 \times V}}{{2V}} = {N_3}$
$ \Rightarrow 0.05 = {N_3}$
Since the normality of the resulting solution is 0.05N thus it behaves as an acid so the pH value of the resulting solution will be less than 7.

So, the correct answer is Option D.

Note: Given that the molarity of sulphuric acid is 0.2 M and the molarity of the potassium hydroxide is 0.1M. And as we know on dissociation sulphuric acid gives 2 hydrogen ions. thus the normality of the sulphuric acid is $0.1\times 2$ = 0.2 N.


Similarly on dissociation potassium hydroxide gives 1 hydroxide ion then the normality of the potassium hydroxide is $0.1\times 1$ = 0.1 N
Since one gram equivalent acid neutralizes the one gram equivalent base thus the sulphuric acid left unneutralized in the resulting solution after mixing is
0.2-0.1=0.1
Hence the resulting solution contains acid thus the PH value for the resulting solution is less than 7.