Question

An aqueous solution contains 4% (w/v) NaOH and 5.3% (w/v) Na2CO3. 40 ml of this solution is titrated with 2.0 M- HCl solution, using phenolphthalein indicator. The volume of HCl solution needed for end point is.
(1)
30 ml
(2)
40 ml
(3)
20 ml
(4)
35 ml
Correct Answer: 1

Answer 1

Given :

Concentration of NaOH in the solution = 4 %

Concentration of $Na_2CO_3$ in the solution  = 5.3%

Volume of  the given solution = 40 ml

Molarity of given HCl solution = 2.0 M

To Find :

Required volume of HCl solution for the end point = ?

Solution :

The balanced chemical equation for this reaction can be written as :

$3HCl + Na_2CO_3 + NaOH \rightarrow 3NaCl + 2H_2O + CO_2$

Since the conc of NaOH is in weight by volume , so we have :

In 100 ml Solution , mass of NaOH =  4  g

So, in 40 ml solution mass of NaOH = $\frac{4}{100} \times 40 = 1.6 g$

So, no of moles of NaOH = $\frac{1.6}{40 } = 0.04$  moles

And , no of moles of HCl = $3 \times$ moles of NaOH

                                         $3 \times 0.04 =0.12$ moles

∴  no of moles =molarity $\times$ volume  

So, volume =$\frac{no \hspace3 of \hspace3 moles \hspace3 of \hspace3 HCl}{molarity \hspace3 of \hspace3 HCl }$

                 = $\frac{0.12}{2} = 0.06 l$

                 $= 0.06 \times 1000 ml = 60 ml$

So, the required volume of HCl for the titration is 60 ml  .