Question #93a69

Question

1719 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-27T08:25:21

Molarity=0.79M,Normalty=0.79N

$"pH of HCl"=-log_10(0.79)=0.1023$

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

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

$"Equivalent mass of NaOH"="molar mass"/"acidity"=40g/"equivalent"$

$"Strength of NaOH"$
$="mass"/"volume in mL"xx(1000mL)/L$

$=(2g)/(100 mL)xx(1000mL)/L=20g/L$

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

Now

$V_a->"Volume of HCl solution"=10mL$

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

$V_b->"Volume of NaOH solution"=15.8mL$

$S_a->"Srength of NaOH solution"=0.5N$

Now applying principle of neutralisation,

$S_axxV_a=S_bxxV_b$

$=>S_axx10=0.5xx15.8$

$=>S_a=0.79N$

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

Molarity of Acid solution:
=0.79M

And the

$"pH of HCl"=-log_10(0.79)=0.1023$