Q.10.98 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the volume, in millilitres, of a 0.215 M KOH solution that will completely neutralize each of the following: 

a. 2.50 mL of a $0.825 \mathrm{M} {\mathrm{H}}_2{\mathrm{SO}}_4$ solution

b. 18.5 mL of a $0.560 \mathrm{M} {\mathrm{HNO}}_3$ solution

c. 5.00 mL of a $3.18 \mathrm{M} \mathrm{HCl}$ solution

Step-by-Step Solution

Answer

Part (a) . $19.1 mL$

Part (b) . $46.9 mL$

Part (c) . $73.9 mL$

Step 1: Information Part (a)

pH or potential of hydrogen is a scale used to specify the acidity or basicity of a fluid solution. Acidic solutions are estimated to have lower pH values than basic or alkaline solution

Step 2: Explanation Part (a)

Given, $0.215 \mathrm{M} \mathrm{KOH}$ solution

Molarity of KOH =$0.215 M$and Molarity of $H_{2}S{O}_4$=$0.825M$

Volume of $H_{2}S{O}_4$= $2.5 \mathrm{ml}=0.0025 \mathrm{L}$

the reaction between these is,

${\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+2\mathrm{KOH}\left(\mathrm{aq}\right)\to {\mathrm{K}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+2{\mathrm{H}}_{2}0\text{ (l) }$

Number of moles of $H_{2}S{O}_4$ = $0.0025\times 0.825=0.0020625\text{ moles }$

From the reaction the moles of KOH = $2\times$(moles of $H_{2}S{O}_4$)

Number of moles of KOH = $2\times 0.0020625=0.004125\text{ moles }$

The volume of KOH = $\frac{number of moles }{molarity}=\frac{0.004125}{0.215}=19.1 mL$

Step 3: Explanation Part (b)

Given, $0.215\mathrm{M} \mathrm{KOH}$solution

Molarity of KOH = $M_2=0.215 M$and Molarity of $HN{O}_3=M_1=0.56 M$

The volume of $HN{O}_3=V_1=18mL$and Volume of  KOH = $V_2$

the reaction between these is,

${\mathrm{HNO}}_3\left(\mathrm{aq}\right)+\mathrm{KOH}\left(\mathrm{aq}\right)\to {\mathrm{KNO}}_3\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\text{ (l) }$

Using, $M_1{V}_1=M_2{V}_2$

${\mathrm{V}}_2=\frac{18\times 0.560}{0.15}=46.9 \mathrm{ml}$

Step 4: Explanation Part (c)

Given, $0.215\mathrm{M} \mathrm{KOH}$ solution

Molarity of HCL = ${\mathrm{M}}_1=3.18\mathrm{M}$ and Molarity of KOH = $M_2=0.215 \mathrm{M}$

The volume of HCL = $V_1=5mL$ and The volume of KOH = $V_2$

Using, $M_1{V}_1=M_2{V}_2$

${\mathrm{V}}_2=\frac{5\times 3.18}{0.215}=73.9 \mathrm{ml}$