Question 93a69

Jul 27, 2016

Molarity=0.79M,Normalty=0.79N

$\text{pH of HCl} = - {\log}_{10} \left(0.79\right) = 0.1023$

Explanation:

It is given that 2g of NaOH is dissolved in water to prepare a 100mL solution.

$\text{Now molar mass of NaOH"=(23+16+1)g/"mol" =40g/"mol}$

$\text{Equivalent mass of NaOH"="molar mass"/"acidity"=40g/"equivalent}$

$\text{Strength of NaOH}$
$= \text{mass"/"volume in mL} \times \frac{1000 m L}{L}$

$= \frac{2 g}{100 m L} \times \frac{1000 m L}{L} = 20 \frac{g}{L}$

"Strength in normality"=(20g/L)/(40g/"equivalent")=0.5N

Now

${V}_{a} \to \text{Volume of HCl solution} = 10 m L$

S_a->"Srength of HCl solution."=?#

${V}_{b} \to \text{Volume of NaOH solution} = 15.8 m L$

${S}_{a} \to \text{Srength of NaOH solution} = 0.5 N$

Now applying principle of neutralisation,

${S}_{a} \times {V}_{a} = {S}_{b} \times {V}_{b}$

$\implies {S}_{a} \times 10 = 0.5 \times 15.8$

$\implies {S}_{a} = 0.79 N$

Hence Normality of Acid solution:
=0.79N
The basicity of HCl being 1

Molarity of Acid solution:
=0.79M

And the

$\text{pH of HCl} = - {\log}_{10} \left(0.79\right) = 0.1023$