91Ó°ÊÓ

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

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

Given values are:

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

Therefore,

$\text{³¾´Ç±ô±ð r²¹³Ù¾±´Ç=}\frac{{\text{1″¾´Ç±ô â¶Ä‰H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{2³¾´Ç±ô â¶Ä‰N²¹°¿±á}}=\frac{1}{2}$

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

$\begin{array}{l}\text{V}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)\text{׳§}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)\text{=V}\left(\text{NaOH}\right)\text{׳§}\left(\text{NaOH}\right)×\frac{\text{1″¾´Ç±ô }\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)\text{ }}{\text{2″¾´Ç±ô}\left(\text{NaOH}\right)}\\ ⇒\text{S}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)\text{=}\frac{\text{V}\left(\text{NaOH}\right)\text{׳§}\left(\text{NaOH}\right)}{\text{V}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)}×\frac{1}{2}\\ ⇒\text{S}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)=\frac{\text{0.1850 M â¶Ä‰Ã—26.25³¾±ô}}{\text{25ml}}×\frac{1}{2}\\ ⇒\text{S}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)=\text{0.097M}\end{array}$