Maintain law, order and equilibrium...

Chemistry Level 5

Find the solubility(in M M ) of A g 3 P O 4 Ag_{3}PO_{4} in acidified solution of p H pH = 4 =4 .

G i v e n : Given :

  • K a 1 Ka_{1} = = 1 0 − 3 10^{-3}

  • K a 2 Ka_{2} = = 1 0 − 4 10^{-4}

  • K a 3 Ka_{3} = = 1 0 − 5 10^{-5}

  • K s p ( A g 3 P O 4 ) K_{sp(Ag_{3}PO_{4})} = = 1 0 − 6 10^{-6}


The answer is 0.090133447.

Richeek Das by Richeek Das , 20, India

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1 solution

Richeek Das
Apr 25, 2019

[ A g + ] 4 [Ag^{+}]^{4} = = 3 K s p 3K_{sp} ( 1 + [ H + ] K a 3 + [ H + ] 2 K a 3 . K a 2 + [ H + ] 3 K a 3 . K a 2 . K a 1 ) (1+\frac{[H^{+}]}{Ka_{3}}+\frac{[H^{+}]^{2}}{Ka_{3}.Ka_{2}}+\frac{[H^{+}]^{3}}{Ka_{3}.Ka_{2}.Ka_{1}})

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How does that come?

Toshit Jain - 1 year, 4 months ago

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