Question

100 mL of 0.1N NaOH is mixed with 100mL of 0.1N ${{H}_{2}}S{{O}_{4}}$ . The pH of the resultant solution is:
A. <7
B. >7
C. =7
D. Cannot be predicted

Answer 1

pH is the measure of the acidity or alkalinity of a solution. The term pH stands for ‘Potential of Hydrogen’. The pH scale varies from 0 to 14. If the pH is more than 7, the solution is basic, if the pH is less than 7, the solution is acidic and if the pH is equal to 7, the solution is equal to 7.

Complete step by step answer:
- In the above question we see that there is a solution of a strong acid and a strong acid. So there is 100% dissociation.
- The name of the reaction is neutralization reaction. We will now use the data provided in the question to find the pH of the resultant solution.
- It is given that, volume of NaOH = 100mL or 0.1L
Normality of NaOH = 0.1N
Here, normality = molarity = 0.1M
- So, we can calculate the number of moles: concentration (in M) volume (in L)
Number of moles = 0.1 0.1 = 0.01mol
- It is given that, volume of ${{H}_{2}}S{{O}_{4}}$ = 100mL or 0.1L
- Normality of ${{H}_{2}}S{{O}_{4}}$ = 0.1N
- The number of electron change (n) for ${{H}_{2}}S{{O}_{4}}$ = 2
- As we know, N = M$\times$ n
where, N = normality
M = molarity
n = no. of electron change
- Thus,

& Molarity=\dfrac{Normality}{n} \\\ & M=\dfrac{0.1}{2}=0.05M \\\ \end{aligned}$$ \- So, the number of moles will be: concentration (in M) volume (in L) Number of moles = 0.05 $\times $ 0.1 = 0.005mol \- Now, 0.01mol of NaOH exactly neutralizes 0.005mol of ${{H}_{2}}S{{O}_{4}}$ according to the following reaction $$NaOH+{{H}_{2}}S{{O}_{4}}\to N{{a}_{2}}S{{O}_{4}}+{{H}_{2}}O$$ \- Thus neither NaOH is in excess nor ${{H}_{2}}S{{O}_{4}}$ \- Therefore, the solution will be neutral and pH = 7 **The correct answer is option “C” .** **Note:** The possibility of making a mistake is that you forgot to consider the electron change in ${{H}_{2}}S{{O}_{4}}$ and multiplying it directly. n-factor: It is defined as the number of $H^{+}$ substances replaced by 1 mol of an acid in a reaction. Here, the n-factor of ${{H}_{2}}S{{O}_{4}}$ = 2 as 2$H^{+}$ will form when it dissociates.