Calculate the molarityes and normalities of the solution obtained on mixing
(i) 100 mL of 0.2 M `H_(2)SO_(4)` with 50 mL of 0.1 M HCl
(ii) 100 mL of 0.2 N `H_(2)SO_(4)` with 50 mL of 0.1 N HCl
(iii) 100 mL of 0.1 M `H_(2)SO_(4)` with 50 mL of 0.1 M NaOH
(iv) 50 mL of 0.1 N `H_(2)SO_(4)` with 100 mL of 0.1 N NaOH.

Explanation

(i) `"100 mL of 0.2 M "H_(2)SO_(4)" contain "H_(2)SO_(4)=(0.2)/(1000)xx100=0.02mol="0.02 g eq."`
`"50 mL of 0.1 M HCl contain HC"=(0.1)/(1000)xx50="0.005 mol = 0.005 g eq."`
`"Total no. of moles present "=0.02+0.005mol=0.025 mol`
`"Total volume after mixing "=100+50mL=150mL=0.150L`
`"Molarity"=(0.025mol)/(0.150 L)=0.167M`
`"Total no. of g eq."=0.04+0.005="0.045g eq."`
`"Normality"=("0.045 g eq")/("0.150 L")=0.3N`
Alternatively, `M_(1)V_(1)+M_(2)V_(2)=M_(3)(V_(1)+V_(2))`
`therefore" "M_(3)=(0.2xx100+0.1xx50)/(150)=0.167M`
Similarly, `N_(1)V_(1)+N_(2)V_(2)=N_(3)(V_(1)+V_(2))`
`"As 0.2 M "H_(2)SO_(4)="0.4 N "H_(2)SO_(4) and " 0.1 M HCl = 0.1 N HCl"`
`N_(3)=(0.4xx100+0.1xx50)/(150)=0.3N`
(ii) `"0.2 N "H_(2)SO_(4)="0.1 M "H_(2)SO_(4)" and 0.1 N HCl = 0.1 M HCl"`
`"Applying "M_(1)V_(1)+M_(2)V_(2)=M_(3)(V_(1)+V_(2))`
`M_(3)=(0.1xx100+0.1xx50)/(150)=(15)/(150)=0.1M`
`"Applying "N_(1)V_(1)+N_(2)V_(2)=N_(3)(V_(1)+V_(2))`
`N_(3)=(0.2xx100+0.1xx50)/(150)=(25)/(150)=0.167N`
(iii) `"100 mL of 0.1 M "H_(2)SO_(4)" contain "H_(2)SO_(4)=(0.1)/(1000)xx100="0.01 mol = 0.02 g eq"`
`"50 mL of 0.1 M NaOH contain NaOH"=(0.1)/(1000)xx50="0.005 mol = 0.005 g eq"`
`"0.005 g eq. of NaOH neutralize 0.005 g eq. of "H_(2)SO_(4)," therefore, "H_(2)SO_(4)" left unneutralized in the solution"`
`=0.02-0.005="0.015 g eq."`
`"Total volume of solution after mixing = 150 mL = 0.150 L"`
`therefore" Normality of "H_(2)SO_(4)" in the solution"=("0.015g eq")/(0.150L)="0.1 N. Hence, molairy"=("Normality")/("Basicity")=(0.1)/(2)="0.05 M"`
(iv) `"50 mL of 0.1 N "H_(2)SO_(4)" contain "H_(2)SO_(4)=(0.1)/(1000)xx"50 g eq. = 0.005 g eq."`
`"100 mL of 0.1 N NaOH contain NaOH"=(0.1)/(1000)xx"100 g eq. = 0.01 g eq."`
`"0.005 g eq. of "H_(2)SO_(4)" neutralize 0.005 g eq. of NaOH, therefore, NaOH left in the solution"`
`=0.01-0.005="0.005 geq."`
`"Total volume of solution after mixing = 150 mL = 0.150 L"`
`therefore" Normality of NaOH in the solution"=("0.005 g eq")/("0.150 L")="0.033 N. Hence,"`
`"molarity x "=("Mormality")/("Acidity")=(0.033)/(1)=0.033M`