Answer:
[tex]m_{PtBr_4}=0.306gPtBr_4[/tex]
Explanation:
Hello,
In this case, since the solubility product of platinum (IV) bromide is 8.21x10⁻⁹, and the dissociation is:
[tex]PtBr_4(s)\rightleftharpoons Pt^{4+}(aq)+4Br^-(aq)[/tex]
The equilibrium expression is:
[tex]Ksp=[Pt^{4+}][Br^-]^4[/tex]
Thus, since the salt is added to a solution initially containing 1.00 grams of potassium bromide, there is an initial concentration of bromide ions:
[tex][Br^-]_0=\frac{1.00gKBr*\frac{1molKBr}{119gKBr}*\frac{1molBr^-}{1molKBr} }{0.250L}=0.0336M[/tex]
Hence, in terms of the molar solubility [tex]x[/tex], we can write:
[tex]8.21x10^{-9}=(x)(0.0336+4x)^4[/tex]
In such a way, solving for [tex]x[/tex], we obtain:
[tex]x=0.00238M[/tex]
Which is the molar solubility of platinum (IV) bromide. Then, since its molar mass is 514.7 g/mol, we can compute the grams that get dissolved in the 250.0-mL solution:
[tex]m_{PtBr_4}=0.00238\frac{molPtBr_4}{1L}*0.250L *\frac{514.7gPtBr_4}{1molPtBr_4} \\\\m_{PtBr_4}=0.306gPtBr_4[/tex]
Best regards.
