Question

Find the purity percentage of ${\mathrm{H}}_2{\mathrm{SO}}_4$with density$1.8$, if $5\mathrm{ml}$ of ${\mathrm{H}}_2{\mathrm{SO}}_4$ is neutralized completely by $84.6\mathrm{ml}$of $2\mathrm{N}$ $\mathrm{NaOH}$ solution.

Answer 1

Titration:

1. Titration is the process of gradual addition of a solution of a known concentration and volume with another solution of unknown concentration until the reaction approaches its neutralization.
2. Titration is used to find the normality of the acid and base titration:
3. Molarity is No. of moles dissolved in $1000\mathrm{ml}$ solution.

Step-1 Calculating the Normality of ${\mathrm{H}}_2{\mathrm{SO}}_4$

N₁ $=$Normality of the Acidic ${\mathrm{H}}_2{\mathrm{SO}}_4$ solution $=$N

V₂ $=$ Volume of the Acidic ${\mathrm{H}}_2{\mathrm{SO}}_4$solution $=$$5\mathrm{ml}$

N₂ $=$ Normality of the basic$\mathrm{NaOH}$ solution $=$ $2\mathrm{N}$

V₂ = Volume of the basic $\mathrm{NaOH}$ solution $=$ $84.6\mathrm{ml}$

N₁V₁$=$N₂V₂

$\mathrm{N}\times 5=84.6\times 2$

$\mathrm{N}=33.84\mathrm{N}$

Step-2 Converting normality into molarity

Molarity$=$Normality$÷$basicity

$=33.84\mathrm{N}÷2\Rightarrow 16.92\mathrm{M}$

$16.92$moles of solutes dissolve in $1000\mathrm{ml}$ of solution to give $16.92\mathrm{M}$solution

Step-3 Calculating the Mass of ${\mathrm{H}}_2{\mathrm{SO}}_4$ using density

1. Given the density of ${\mathrm{H}}_2{\mathrm{SO}}_4$$=$$1.8$
2. Volume of ${\mathrm{H}}_2{\mathrm{SO}}_4$ from molarity$=$$1000mL$
3. $\mathrm{Density}=\frac{\mathrm{Mass}}{\mathrm{Volume}}$,
4. $\mathrm{Mass}=\mathrm{Density}\times \mathrm{Volume}$
5. The mass of ${\mathrm{H}}_2{\mathrm{SO}}_4$$=$Number of moles of ${\mathrm{H}}_2{\mathrm{SO}}_4$ $\times$Molecular weight$\Rightarrow$$16.92\times 98\Rightarrow 1658.16\mathrm{g}$

Step-4 Calculating the Percentage purity of ${\mathrm{H}}_2{\mathrm{SO}}_4$

Percentage Purity$=$(Mass of $16.92$ moles of ${\mathrm{H}}_2{\mathrm{SO}}_4$$÷$Mass of ${\mathrm{H}}_2{\mathrm{SO}}_4$ with density $1.8$)$\times 100$

$=$$(1658.16÷1800)\times 100\Rightarrow 92.12\%$

Hence, the purity percentage of ${\mathrm{H}}_2{\mathrm{SO}}_4$with density $1.8$= $92.12\%$