91Ó°ÊÓ

Whenever there is a reaction in a fluid arrangement, the water atoms can draw in and briefly hold a given proton $H^{+}$. This makes the hydronium particle $H_3{O}^{+}$. In an acidic watery arrangement, the centralization of hydronium particles will be higher than the grouping of hydroxide $O{H}^{-}$particles.

We know,

$0.05M$KOH solution

Using,

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{−}\right]=1.0×{10}^{−14}$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\frac{1.0×{10}^{−14}}{0.050\mathrm{M}}=2×{10}^{-13}M$

We know,

$0.05M$KOH solution

Using,

$\mathrm{pH}=−\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

$$\begin{gathered} \mathrm{pH}=−\log \left[2.0×{10}^{−13}\right] \\ =12.7 \end{gathered}$$

The balanced chemical reaction is,

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

Given,

Volume = $40ml$of a $0.035\mathrm{M}{\mathrm{H}}_2{\mathrm{SO}}_4$solution.

We know,$1\mathrm{mol}{\mathrm{H}}_2{\mathrm{SO}}_4=2\mathrm{mol}\mathrm{KOH}$from the reaction.

$1\mathrm{L}\mathrm{KOH}=0.050\mathrm{mol}\mathrm{KOH}$

Now calculating,

$$\begin{gathered} \frac{40m{\mathrm{L}\mathrm{H}}_2{\mathrm{SO}}_4}{1000\mathrm{mL}{\mathrm{H}}_2{\mathrm{SO}}_4}×\frac{0.035\mathrm{mol}{\mathrm{H}}_2{\mathrm{SO}}_4}{1{\mathrm{L}\mathrm{H}}_2{\mathrm{SO}}_4}×\frac{2\mathrm{mol}\mathrm{KOH}}{1\mathrm{mol}{\mathrm{H}}_2{\mathrm{SO}}_4}×\frac{1\mathrm{L}\mathrm{KOH}}{0.050\mathrm{mol}\mathrm{KOH}}×\frac{1000\mathrm{mL}\mathrm{KOH}}{1L\mathrm{KOH}} \\ =56mLKOH \end{gathered}$$