Question

A chemistry student weighs out of sulfurous acid , a diprotic acid, into a volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with solution. A.Calculate the volume of solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.

Answer 1

Answer: 36.2 ml

Explanation:

Assmed values :

weight of sulphurous acid = 0.104 g

volume of flask = 250 ml

Molarity of NaOH solution = 0.0700 M

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute  = $\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.104g}{82.07g/mol}=0.00127mole$  

$V_s$ = volume of solution in ml

$Molarity=\frac{0.00217\times 1000}{250ml}=0.00507M$

According to the neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$M_1$ = molarity of $H_2SO_3$ solution = 0.00507 M

$V_1$ = volume of $H_2SO_3$ solution = 250 ml

$M_2$ = molarity of $NaOH$ solution = 0.0700 M

$V_2$ = volume of $NaOH$ solution = ?

$n_1$ = valency of $H_2SO_3$ = 2

$n_2$ = valency of $NaOH$ = 1

$2\times 0.00507M\times 250=1\times 0.0700\times V_2$

$V_2=36.2ml$

Therefore, the volume of solution the student will need to add to reach the final equivalence point is 36.2 ml