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#