Calculate the normality of mixture obtained by mixing
a. `100 mL of 0.1 N HCl + 50 mL of 0.25 N NaOH`
b. `100 mL of 0.2 M H_(2) SO_(4) + 200 mL of 0.2 M HCl`
c. `100 mL of 0.2 M H_(2) SO_(4) + 100 mL of 0.2 M NaOH`
d. `1g` equivalent of `NaOH + 100 mL of 0.1 N HCl`

Explanation

(a) Meq. of `H_(2)SO_(4)=100xx0.1=10`
Meq. of `NaOH=50xx0.25=12.5`
`H_(2)SO_(4)` and `NaOH` neutralize each other.
Meq. of `NaOH` left `=12.5-10=2.5`
Volume of new solution`=100+50=150 mL`
`:. N_(NaOH) "left" =2.5//150=0.0167`
(b) Meq. Of `H_(2)SO_(4)=100xx0.2xx2=40, ( :' N=MxxValancy)`
Meq. of `HCl =200xx0.2xx1=40`
`:.`total Meq. of acid`=40+40=80`
`:.` Total volume of solution `=300 mL`
`:. N_(Acid solution)=80//300=0.267`
(c ) Meq. of `H_(2)SO_(4)=100xx0.2xx2=40`
Meq. of `NaOH=100xx0.2xx1=20`
Meq. of `H_(2)SO_(4)` left after reaction`=40-20=20`
Total volume of solution `=100+100=200 mL`
`:. N_(H_(2)SO_(4)) "left" =20//200=0.1`
(d) Meq. of `NaOH=1xx1000=1000`
Meq. of `HCl=100xx0.1=10`
Meq. of `NaOH` left after reaction `=1000-10=990`
Total volume of solution `=100mL`
`N_(NaOH) "left"=990//100=9.9`