Two acids $H_{2} {S O}_{4}$ and $H_{3} {P O}_{4}$ are neutralized separately by the same amount of an alkali when sulphate and dihydrogen phosphates are formed respectively.Find the ratio of the masses of $H_{2} {S O}_{4}$ and $H_{3} {P O}_{4}$

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Solution

$H_{2} S O_{4} + 2 N a O H ⟶ N a_{2} S O_{4} + 2 H_{2} O$
$H_{3} P O_{4} + N a O H ⟶ N a H_{2} P O_{4} + H_{2} O$
Equivalent of alkali= $1 g$ equivalent of $H_{2} S O_{4} = 1 g$ equivalent of $H_{3} P O_{4}$
Two acids should react in their equivalent mass.
Equivalent mass of $H_{2} S O_{4} = \frac{2 + 32 + 64}{2}$
$= \frac{98}{2} = 49$
Equivalent mass of $H_{3} P O_{4} = \frac{3 + 31 + 64}{1}$
$= \frac{98}{1} = 98$
ratio of equivalent masses
$= 49 : 98$
$= 1 : 2$

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H_{2} S O_{4}

H_{2} S O_{4}

H_{3} P O_{4}

H_{2} {P O}_{4}^{-}

H_{3} {P O}_{4} + H_{2} O \to H_{3} O^{+} + H_{2} {P O}_{4}^{-}

H_{2} {P O}_{4}^{-} + H_{2} O \to H {P O}_{4}^{2 -} + H_{3} O^{+}

H_{2} {P O}_{4}^{-} + {O H}^{-} \to H_{3} {P O}_{4} + O^{2 -}

H_{2} O + H_{3} {P O}_{4} ⇌ H_{3} O^{+} + H_{2} {P O}_{4}^{-}; p K_{1} = 2.15

H_{2} O + H_{2} {P O}_{4}^{-} ⇌ H_{3} O^{+} + H_{2} {P O}_{4}^{2 -}; p K_{2} = 7.20

0.01 M

N a H_{2} {P O}_{4}

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