Question

The K sp for silver(I) phosphate is 1.8 × 10 –18. Determine the silver ion concentration in a saturated solution of silver(I) phosphate.

Answer 1

`[Ag^+]=4.82x10^(-5)M`

Step-by-step explanation:

Hello,

In this case, the dissociation reaction for silver phosphate is:

`Ag_3PO_4(s)\rightleftharpoons 3Ag^+(aq)+PO_4^(3-)(aq)`

Therefore, the equilibrium expression is:

`Ksp=[Ag^+]^3[PO_4^(3-)]`

And in terms of the reaction extent
`x` is:

`Ksp=1.8x10^(-18)=(3x)^3(x)`

Thus,
`x` turns out:

`1.8x10^(-18)=27x^4\\\\x=\sqrt[4]{(1.8x10^(-18))/(27) } \\\\x=1.61x10^(-5)M`

In such a way, the concentration of the silver ion is:

`[Ag^+]=3x=3*1.61x10^(-5)M=4.82x10^(-5)M`

Best regards.