Respuesta :
The given question is incomplete. The complete question is as follows.
A basic solution contains the iodide and phosphate ions that are to be separated via selective precipitation. the i– concentration, which is 9.00×10-5 m, is 10,000 times less than that of the [tex]PO^{3-}_{4}[/tex] ion at 0.900 m . a solution containing the silver(i) ion is slowly added. answer the questions below. ksp of agi is 8.30×10-17 and of [tex]Ag_{3}PO_{4}[/tex], [tex]8.90 \times 10^{-17}[/tex].
Calculate the minimum [tex]Ag^{+}[/tex] concentration required to cause precipitation of AgI.
Explanation:
It is known that at the stage of precipitation, the [tex]K_{sp}[/tex] is equal to the ionic product.
Therefore, expression for [tex]K_{sp}[/tex] for [tex]Q_{1}[/tex] is as follows.
[tex]K_{sp} = [Ag^{+}][I^{-}][/tex]
In the given case,
[tex][I^{-}] = 7.7 \times 10^{-5}[/tex]
[tex]K_{sp} = 8.3 \times 10^{-17}[/tex]
Hence, [tex][Ag^{+}] = \frac{k_{sp}}{I^{-}}[/tex]
= [tex]\frac{8.3 \times 10^{-17}}{9 \times 10^{-5}}[/tex]
= [tex]0.92 \times 10^{-12} M[/tex]
Now, for [tex]Q_{2}[/tex] the expression for [tex]K_{sp}[/tex] will be as follows.
[tex]K_{sp} = [Ag^{+}]^{3}[PO^{3-}_{4}][/tex]
or, [tex][Ag^{+}] = (\frac{K_{sp}}{[PO^{3-}_{4}]})^{\frac{1}{3}}[/tex]
= [tex](\frac{8.90 \times 10^{-17}}{0.9 \times 10^{-5}})^{\frac{1}{3}}[/tex]
= [tex]9.72 \times 10^{-5}[/tex] M
