Question

In the titration of $\mathbf{100}\mathbf{.}\mathbf{0}\mathbf{mL}\mathbf{of}\mathbf{a}\mathbf{0}\mathbf{.}\mathbf{0500}\mathbf{M}$ solution of acid${\mathbf{H}}_3\mathbf{A}$ calculate the volume of$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{M}\mathbf{NaOH}$ required to reach pH values of$\mathbf{9}\mathbf{.}\mathbf{50}\mathbf{and}\mathbf{4}\mathbf{.}\mathbf{00}$

Short answer

In order to reach$\text{pH}=\text{9.50}$ required volume of$\text{NaOH}$ is$10.0\text{\hspace{0.17em}mL}$

In order to reach$\text{pH}=4.\text{00}$ required volume of$\text{NaOH}$ is$\text{4.55\hspace{0.17em}mL}$

Step-by-step solution

Definition of volume

The amount of space occupied by a substance, while mass is the amount of matter it contains is defined as volume.

Calculation of volume

$\begin{array}{l}\text{V}=\frac{\text{n}}{\text{M}}\\ \text{V}=\frac{10\times {10}^{-3}\text{mol}}{1.0\text{\hspace{0.17em}M}}\\ \text{V}=10\text{\hspace{0.17em}mL}\end{array}$

In order to reach $\text{pH}=\text{9.50}$required volume of $\mathrm{NaOH}$ is $10.0\text{\hspace{0.17em}mL}$

The number of moles volume can be calculated as

$\begin{array}{c}\text{V=}\frac{\text{n}}{\text{M}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=}\frac{{\text{4.55×10}}^{\text{-3}}\text{\hspace{0.17em}mol}}{\text{1\hspace{0.17em}M}}\\ \text{V=4.55\hspace{0.17em}mL}\end{array}$

In order to reach$\text{pH}=4.\text{00}$ required volume of$\text{NaOH}$ is$\text{4.55\hspace{0.17em}mL}$.