What will be the pH of a $0.01 M H_{3} P O_{4}$ solution having $[P O_{4}^{3 -}] = 10^{- 5} M$
$[K_{a_{1}} = 10^{- 4}, K_{a_{2}} = 10^{- 6}, K_{a 3} = 10^{- 8}]$

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answer is 5.

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Detailed Solution

$K_{a} = 10^{- 4} \times 10^{- 6} \times 10^{- 8} = 10^{- 18} o r, 10^{- 18} = \frac{{[H^{+}]}^{3} [P O_{4}^{3 -}]}{[H_{3} P O_{4}]} \Rightarrow [H^{+}] = 10^{- 5} \Rightarrow p H = 5$

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