Question

4g of naoh is weighed and made upto 500 ml . The volume of 0.1N Hcl required to neutralise completly 10 ml of this solution

Answer 1

Given:

4g of NaOH is weighed and made upto 500 ml.

To find:

Volume of 0.1N Hcl required to neutralise completely 10 ml of this solution?

Calculation:

First of all, let's find the normality of NaOH solution:

$\rm normality = \dfrac{molarity}{n \: factor}$

$\rm \implies normality = \dfrac{( \frac{moles}{volume}) }{n \: factor}$

$\rm \implies normality = \dfrac{( \frac{0.1}{500} \times 1000) }{1}$

$\rm \implies normality = 0.2 \: N$

Now, applying titration expression:

$\rm \: V_{1}S_{1} = V_{2}S_{2}$

$\rm \implies\:10 \times 0.2= V_{2} \times 0.1$

$\rm \implies\: V_{2} = 20 \: ml$

So, volume of HCl required is 20 mL.

Answer 2

Answer: 20 mL of 0.1N Hcl is required to neutralise completly 10 ml of this solution