$50 m L$ of $0.12 M$ $H_{3} {P O}_{4} + 20 m L$ of $0.15 M$ $N a O H$

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Solution

For $H_{3} P O_{4}$ , $K_{a 1} >> K_{a 2} >> K_{a 3}$ . Hence, we will consider it as monoprotic acid.
$p K_{a} = - l o g_{10} K_{a 1} = - l o g_{10} 7.5 \times 10^{- 3} = 2.12 ≃ 2$
The number of mmol of $H_{3} P O_{4} = 50 m L \times 0.12 M = 6 m m o l$
The number of mmol of $N a O H = 20 m L \times 0.15 M = 3 m m o l$
3 mmol of $N a O H$ will neutralise 3 mmol of $H_{3} P O_{4}$
Out of 6 mmol of $H_{3} P O_{4}$ , 3 mmol are neutralised. This correspond to half-equivalence point.
At half-equivalence point, $p H = p K_{a 1} = 2$

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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 -}

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