Question

What is the resultant ${pH}$ of $0.1M$ ${H_2}S{O_4}$ of $100ml$ and $0.3N$ of $NaOH$ of $100ml$.

Answer 1

The normality of sulphuric acid is the double of the molarity as the valence factor of ${H_2}S{O_4}$ is $2$. The Concentration of hydroxide ion concentration can be obtained from the formula. The ${p{OH}}$ can be calculated from the hydroxide ion concentration. The ${pH}$ can be calculated from ${p{OH}}$.
Formula used:
$\left[ {O{H^ - }} \right] = \dfrac{{{N_2}{V_2} - {N_1}{V_1}}}{{{V_1} + {V_2}}}$
$\left[ {O{H^ - }} \right]$ is concentration of hydroxide ion
${N_2}$ is normality of $NaOH$ is $0.3N$
${V_2}$ is volume of $NaOH$ is $0.1L$
${N_1}$ is normality of ${H_2}S{O_4}$ is $0.1 \times 2 = 0.2N$
${V_1}$ is volume of ${H_2}S{O_4}$ is $0.1L$

Complete answer: Given that the molarity of ${H_2}S{O_4}$ is $0.1M$ . As the valence factor of sulphuric acid is two. The normality will be double to the molarity. Thus, the normality of ${H_2}S{O_4}$ is $0.2N$.
Substitute all the values in the above formula,
$\left[ {O{H^ - }} \right] = \dfrac{{0.3\left( {0.1} \right) - 0.2\left( {0.1} \right)}}{{0.1 + 0.1}}$
Upon simplification of the above equation, we will get
$\left[ {O{H^ - }} \right] = \dfrac{{0.01}}{{0.2}} = 0.05$
Thus is the concentration of hydroxide ions.
${pH}$ is defined as the negative logarithm of the $\left[ {{H^ + }} \right]$ ion concentration.
The ${p{OH}}$ will be obtained by substituting the concentration of hydroxide ion in ${p{OH}} = - \log \left( {O{H^ - }} \right)$
Thus, ${p{OH}} = - \log \left( {0.05} \right) = 1.3$
The equation related to ${pH}$ and ${p{OH}}$ is ${pH} + {p{OH}} = 14$
The acids have the ${pH}$ below $7$ and bases have ${pH}$ above $7$.
The value of ${pH}$ will be obtained by substituting the value of ${p{OH}}$ in the above equation.
${pH} = 14 - 1.3 = 12.7$
Thus, the resultant ${pH}$ of $0.1M$ ${H_2}S{O_4}$ of $100ml$ and $0.3N$ of $NaOH$of $100ml$ is $12.7$.

Note:
The normality and molarity are the units used to express the concentration of a substance. But the normality for acids is molarity by valence factor and for bases it is molarity by valence factor. The normality for salts will be molarity by number of electrons oxidized or reduced. So, the normality must be calculated accurately.