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#
