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 `HCl = 100 xx 0.25 = 10`
mEq of `NaOH = 50 xx 0.25 = 12.5`
Because `HCl` and `NaOH` neutralise each other with equal equivalents
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) = 100 xx 0.2 xx 2 ( :' N = M xx` Valency)
mEq of `HCl = 200 xx 0.2 xx 1 = 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) = 100 xx 0.2 xx 2 = 40`
mEq of `NaOH = 100 xx 0.2 xx 1 = 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 = 1 xx 1000 = 1000`
mEq of `HCl = 100 xx 0.1 = 10`
`:.` mEq of `NaOH` after reaction `= 1000 - 10 = 990`
Total volume of solution `= 100 mL`
`:. N_(NaOH)` left `= (990)/(100) = 9.9`