A metal weighing 0.43 g was dissolved in 50 mL of 1 N `H_(2)SO_(4)`. The unreacted `H_(2)SO_(4)` required 14.2 mL of 1 N NaOH for neutralisation. Calculate the equivalent weight of the metal?

Explanation

`H_(2)SO_(4)" left unreacted "-=NaOH" reacted = 14.2 mL of 1 N"`
`therefore" 1 N "H_(2)SO_(4)" reacted with the metal "=(50-14.2)mL=35.8mL`
`"35.8 mL of 1 N "H_(2)SO_(4)="35.8 milliequivalents"`
If E is the equivalent wt. of the metal, milliequivalents of the metal reacted `"=(0.43)/(E)xx1000=(430)/(E)`
As metal and the acid will react in equivalent amounts,
`(430)/(E)=35.8 or E=(430)/(35.8)=12.01`