Question

the two acids $H_{2}S{O}_4$ are neutralized separately by the same amount of an alkali when sulphate and dihydrogen orthophate are formed respectively. Find the ratio of the masses of $H_{2}S{O}_4$ and $H_{3}P{O}_4$.

Answer 1

$H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O$
$H_{3}P{O}_4+NaOH\to Na{H}_{2}P{O}_4+H_{2}O$

Equivalent of alkali = 1 g equivalent of $H_{2}S{O}_4$= 1 g equivalent of $H_{3}P{O}_4$

The two acids must react in the ratio of their equivalent masses.

Equivalent mass of $H_{2}S{O}_4=\frac{98}{2}=49$

Equivalent mass of $H_{3}P{O}_4=\frac{98}{1}=98$

Hence ratio of masses of $H_{2}S{O}_4$: $H_{3}P{O}_4$= 49 : 98 =1 : 2.