$9.8 \times 10^{- 7} g H_{3} P O_{4}$ is dissolved in 100mL of a buffer solution of $p H = 5$ (Provides constant concentration of $H^{+}$ ) If the equilibrium concentrations of $H_{3} P O_{4}, H_{2} P O_{4}^{-}, H P O_{4}^{2 -}$ and $P O_{4}^{- 3}$ are $C_{1}, C_{2}, C_{3}$ and $C_{4}$ respectively. Then $P^{C_{1}} + P^{C_{2}} + P^{C_{3}} + P^{C_{4}} =$ _________(Round off to nearest integer)
$(p C_{i} = - \log C_{i}; w h e r e i = 1, 2, 3, 4)$ Given: For $H_{3} P O_{4}, K_{a_{1}} = 10^{- 3}, K_{a_{2}} = 10^{- 8}, K_{a_{3}} = 10^{- 12}$
Since $k a_{1} > > k a_{2}$ and $k a_{3}$
using the approximation
${[H_{3} P O_{4}]}_{0} = [H_{3} P O_{4}] + [H_{2} P {O_{4}}^{-}] {[H_{3} P O_{4}]}_{0} = I n i t i a l c o n c e n t r a t i o n o f a c i d$

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answer is 43.

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Detailed Solution

${[H_{3} P O_{4}]}_{0} = 10^{- 7}$

$a) K_{a 1} = \frac{10^{- 5} [H_{2} P O_{4}^{-}]}{[H_{3} P O_{4}]} = 10^{- 3} [H_{2} P O_{4}^{-}] = 10^{2} [H_{3} P O_{4}] B u t {[H_{3} P O_{4}]}_{o} = [H_{2} P O_{4}^{-}] + [H_{3} P O_{4}] 10^{- 7} = 10^{2} [H_{3} P O_{4}] + [H_{3} P O_{4}] ≃ 10^{2} [H_{3} P O_{4}] C_{1} = [H_{3} P O_{4}] = \frac{10^{- 7}}{(10^{2} + 1)} ≃ 10^{- 9} M$

b) $C_{2} = [H_{2} P O_{4}^{-}] = 10^{2} \times 10^{- 9} = 10^{- 7} M$

$c) K a_{2} = 10^{- 5} \frac{[H P O_{4}^{2 -}]}{[H_{2} P O_{4}^{-}]} = 10^{- 8}$

$C_{3} = [H P O_{4}^{2 -}] = 10^{- 3} \times 10^{- 7} = 10^{- 10}$

$d) K a_{3} = 10^{- 5} \frac{[P O_{4}^{3 -}]}{[H P O_{4}^{2 -}]} = 10^{- 12} C_{4} = [P O_{4}^{3 -}] = 10^{- 7} \times 10^{- 10} = 10^{- 17} P^{C_{1}} + P^{C_{2}} + P^{C_{3}} + P^{C_{4}} = 9 + 8 + 10 + 17 = 44$