Question

Sodium hydroxide is used extensively in acid-base titrations because it is a strong, inexpensive base. A sodium hydroxide solution was standardized by titrating 25.00 mL of 0.1528 M standard hydrochloric acid. The initial buret reading of the sodium hydroxide was 2.24 mL, and the final reading was 39.21 mL. What was the molarity of the base solution?

Short answer

The molarity of the base solution is 0.103 M NaOH.

Step-by-step solution

first write down the balanced molecular equation:

$\mathit{N}\mathit{a}\mathit{O}\mathit{H}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathit{H}\mathit{C}\mathit{l}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\to }\mathit{N}\mathit{a}\mathit{C}\mathit{l}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{l}\mathbf{\right)}$

To find out the molarity of the base solution, first, we have to calculate the moles of sodium hydroxide:

Given values are:

$$\begin{gathered} volumeofstandard0.1528MHClsolution=25ml \\ initialburetereading=2.24ml \\ finalburetereading=39.21ml \end{gathered}$$

Moles ofSodium hydroxide:

$\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathit{o}\mathit{f}\mathit{N}\mathit{a}\mathit{O}\mathit{H}\mathbf{=}\mathbf{\left(}\mathbf{25}\mathit{m}\mathit{l}{\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{1}\mathbf{l}\mathbf{i}\mathbf{t}}{\mathbf{1000}\mathbf{m}\mathbf{l}}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{1528}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{H}\mathbf{C}\mathbf{l}}{\mathbf{1}\mathbf{l}\mathbf{i}\mathbf{t}\mathbf{H}\mathbf{C}\mathbf{l}\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{N}\mathbf{a}\mathbf{O}\mathbf{H}}{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{H}\mathbf{C}\mathbf{l}}\mathbf{\right)}$

$\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{82}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{m}\mathit{o}\mathit{l}\mathit{N}\mathit{a}\mathit{O}\mathit{H}$

The molarity of the base solution is:

$\mathit{m}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathit{o}\mathit{f}\mathit{N}\mathit{a}\mathit{O}\mathit{H}\mathbf{=}\frac{\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{s}\mathbf{o}\mathbf{f}\mathbf{N}\mathbf{a}\mathbf{O}\mathbf{H}}{\mathbf{v}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{m}\mathbf{e}\mathbf{o}\mathbf{f}\mathbf{N}\mathbf{a}\mathbf{O}\mathbf{H}}$

$\mathbf{=}\frac{\mathbf{3}\mathbf{.}\mathbf{82}\mathbf{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{N}\mathbf{a}\mathbf{O}\mathbf{H}}{\mathbf{(}\mathbf{39}\mathbf{.}\mathbf{21}\mathbf{m}\mathbf{l}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{24}\mathbf{m}\mathbf{l}\mathbf{)}\mathbf{\left(}\frac{\mathbf{1}\mathbf{l}\mathbf{i}\mathbf{t}}{\mathbf{1000}\mathbf{m}\mathbf{l}}\mathbf{\right)}}$

$\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{103}\mathit{M}\mathit{N}\mathit{a}\mathit{O}\mathit{H}$