Q.10.107 - An Introduction to General, Organic, and Biological Chemistry

Question

Determine each of the following for a 0.050 M KOH solution: 

a. $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

b. pH

c. the balanced chemical equation for the reaction with ${\mathrm{H}}_2{\mathrm{SO}}_4$

d. millilitres of the KOH solution required to neutralize 40.0 ml of a $0.035 M H_2{\mathrm{SO}}_4$ solution

Step-by-Step Solution

Answer

Part (a) . $2.0\times {10}^{-13} \mathrm{M}$

Part (b) . $12.7$

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

Part (d) .$56 ml$of the KOH solution required

Step 1: Given Information Part (a)

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.

Step 2: Explanation Part (a)

We know,

$0.05 M$ KOH solution

Using, 

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

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

Step 3: Explanation Part (b)

We know,

$0.05 M$KOH solution

Using,

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

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

Step 4: Explanation Part (c)

The balanced chemical reaction is,

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

Step 5: Explanation Part (d)

Given,

Volume = $40 ml$ 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,

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