Q4.54P - Chemistry: Molecular Nature Of Matter And Change

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?

Step-by-Step Solution

Answer

The molarity of the base solution is 0.103 M NaOH.

Step 1: first write down the balanced molecular equation:

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

Step 2: 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:

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

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

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