# Question 76d46

Oct 8, 2016

Let

${V}_{H C l} m L \to \text{Volume of HCl solutoin}$

${S}_{H C l} M \to \text{Strength of HCl solutoin}$

${V}_{A l {\left(O H\right)}_{3}} m L \to \text{Volume of "Al(OH)_3" solutoin}$

Amount HCl taken$= {V}_{H C l} \cdot {S}_{H C l} m m o l$

Amount of $A l {\left(O H\right)}_{3}$ taken

$= {V}_{A l {\left(O H\right)}_{3}} \times 0.22 \text{ } m m o l$

The reaction

$A l {\left(O H\right)}_{3} + 3 H C l \to A l C {l}_{3} + 3 {H}_{2} O$

This equation reveals that the reacting ratio of no. of moles is

Al(OH)_3):HCl= 1:3#

So the residual amount of $A l {\left(O H\right)}_{3}$ is

$= \left({V}_{A l {\left(O H\right)}_{3}} \times 0.22 - \frac{1}{3} \times {V}_{H C l} \cdot {S}_{H C l}\right) m m o l$

The amount of ${H}_{2} S {O}_{4}$ required for neutralisation of ecess $A l {\left(O H\right)}_{3}$ is 5mL 0.38 M
$\equiv 5 \times 0.38 m m o l$
Equation of neutralisation of excess $A l {\left(O H\right)}_{3}$ by ${H}_{2} S {O}_{4}$

$2 A l {\left(O H\right)}_{3} + 3 {H}_{2} S {O}_{4} \to A {l}_{2} {\left(S {O}_{4}\right)}_{3} + 6 {H}_{2} O$

This equation reveals that molar reacting ratio is

$A l {\left(O H\right)}_{3} : {H}_{2} S {O}_{4} = 2 : 3$

So

${V}_{A l {\left(O H\right)}_{3}} \times 0.22 - \frac{1}{3} \times {V}_{H C l} \cdot {S}_{H C l} = \frac{2}{3} \times 5 \times 0.38$

Using this equation we can find out ${S}_{H C l}$ if the value of ${V}_{H C l}$ of and ${V}_{A l {\left(O H\right)}_{3}}$ are known.