Q4.52P - Chemistry: Molecular Nature Of Matter And Change

Question

If 26.25 mL of a standard 0.1850 M NaOH solution is required to neutralize 25.00 mL of ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{SO}}}_{\mathbf{\text{4}}}\mathbf{\text{,}}$ what is the molarity of the acid solution?

Step-by-Step Solution

Answer

The molarity of the acid solution is 0.097 M.

Step 1: Write the balanced molecular equation

The reaction of sodium hydroxide and sulfuric acid is shown below.

$2\mathrm{NaOH}\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to {\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Step 2: Calculate the molarity of sulfuric acid

Given values are: 

$\begin{array}{c}\mathrm{volume} \mathrm{of} \mathrm{NaOH} \mathrm{solution}=\mathrm{V}\left(\text{NaOH}\right)=26.25\mathrm{ml}\\ \mathrm{Concentration} \mathrm{of} \mathrm{NaOH} \mathrm{solution}=\mathrm{S}\left(\text{NaOH}\right)=0.1850\mathrm{M}\\ \mathrm{volume} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4=\mathrm{V}\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)=25\mathrm{ml}\\ \mathrm{Concentration} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4 \mathrm{solution}=\mathrm{S}\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)=?\end{array}$

From the balanced chemical reaction, it is seen that one mole of sulfuric acid neutralizes two moles of sodium hydroxide.

Therefore,

 $\text{mole \hspace{0.17em}ratio =}\frac{{\text{1\hspace{0.17em}mol\hspace{0.17em}\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{2mol\hspace{0.17em}\hspace{0.17em}NaOH}}=\frac{1}{2}$

Using the following relation we can calculate the molarity of the acid solution.

$\begin{array}{l}\text{V}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)\text{×S}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)\text{=V}\left(\text{NaOH}\right)\text{×S}\left(\text{NaOH}\right)\times \frac{\text{1\hspace{0.17em}mol\hspace{0.17em}}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)}{\text{2\hspace{0.17em}mol}\left(\text{NaOH}\right)}\\ \Rightarrow \text{S}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)\text{=}\frac{\text{V}\left(\text{NaOH}\right)\text{×S}\left(\text{NaOH}\right)}{\text{V}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)}\times \frac{1}{2}\\ \Rightarrow \text{S}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)=\frac{\text{0.1850\hspace{0.17em}M\hspace{0.17em}\hspace{0.17em}× 26.25ml}}{\text{25ml}}\times \frac{1}{2}\\ \Rightarrow \text{S}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)=\text{0.097 M}\end{array}$