Chemistry, 25.07.2019 23:00 rohan13

Given the solubility, calculate the solubility product constant (ksp) of each salt at 25°c: (a) pbcro4, s = 4.0 × 10−5 g/l; (b) bac2o4, s = 0.29 g/l; (c) mnco3, s = 4.2 × 10−6 g/l.

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a) PbCrO4:

according to the equation:

PbCrO4(s) → Pb2+(aq) + CrO42-(aq)

so, Ksp = [Pb2+][CrO42-]

by assuming [Pb2+] = [CrO42-] = X

and when S (the solubility) = 4 x 10^-5 g/L

we have first to convert solubility from g/L to mol/L by getting the molar mass of the salt

solubility mol/L = solubility g/L / molar mass of salt

= 4 x 10^-5g/L / 323.2 g/mol

= 1.24 x 10^-7 mol / L

by substitution in Ksp formula:

∴Ksp = X* X

= (1.24 x 10^-7)^2

= 1.54 x 10^-14

b) BaC2O4:

according to this equation:

BaC2O4(s)→ Ba 2+(aq) + C2O4 2- (aq)

So Ksp = [Ba2+][C2O42-]

Assume that [Ba2+] = [C2O42-] = X

when the solubility S = 0.29 g/L = X , so we need to convert S from g/L to
mol / L

solubility mol / L= solubility g/L / molar mass of salt

= 0.29 g/L / 225.34 g/mol

= 0.0013 mol/L

by substitution in ksp formula:

∴Ksp = X^2

= (0.0013)^2

= 1.69 x 10^-6

C) MnCO3:

according to this equation :

MnCO3(s)→ Mn2+(aq) + CO3 2-(aq)

so, Ksp = [Mn2+][CO32-]

assume [Mn2+] = [CO32-] = X

when the solubility s = 4.2 x 10^-6 g/L so we need to convert S from g/L to mol/L by dividing on molar mass.

the solubility mol/L = solubility g/L / molar mass g/mol

=.4.2 x 10^-6 /114.9

= 3.7 x 10^-8 mol/L

by substitution on ksp formula:

∴ Ksp = X*X

= (3.7 x 10^-8)^2

= 1.369 x 10^-15

Answer from: Quest

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Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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Given the solubility, calculate the solubility product constant (ksp) of each salt at 25°c: (a) pbc...