Barium sulfate, BaSO_4, is used in medical imaging of the gastrointestinal tract

Question

Barium sulfate, BaSO_4, is used in medical imaging of the gastrointestinal tract because it is opaque to X rays. A barium sulfate solution, sometimes called a cocktail, is ingested by the patient, whose stomach and intestines can then be visualized via X-ray imaging. If a patient ingests 240 mL of a saturated barium sulfate solution, how much toxic Ba^2+ ion has the patient consumed? The solubility product K_sp of BaSO_4 is 1.10 times 10^-10. Express your answer to three significant figures and include the appropriate units. Part B There are some data that suggest that zinc lozenges can significantly shorten the duration of a cold. If the solubility of zinc acetate, Zn(CH_3COO)_2, is 43.0 g/L, what is the solubility product K_sp of this compound? Express your answer numerically.

Explanation / Answer

A.

We must identify the amoun ot MASS of Ba+2 ions, this can be done with stoichiometric relationships using the solubility constant

By definition

BaSO4(s) <-> Ba+2(aq) + SO4-2(aq)

Note that ratio is 1:1

and the Ksp (equloibrium constant)

Ksp = [Ba+2][SO4-2]

and form experimental data Ksp = 1.1*10^-10

then

Assume that "S" is the solubliity of BaSO4

then, by stoichiometry (1:1 ratio)

[Ba+2] = S

[SO4-2] = S

then

Ksp = S*S

1.1*10^-10 = S*S

S^2 = 1.1*10^-10

S = sqrt(1.1*10^-10 ) = 0.00001048808 or 1.048*10^-5

since S = [Ba+2] also

then

[Ba+2] =  1.048*10^-5 M

we need mass so...

M = mol/V

mol = MV

V = 240 mL = 0.24 L

mol =  (1.048*10^-5)(0.240) = 0.0000025152 mol of Ba+2

we need to change to mass so

mass = mol*MW

MW of Ba = 137.327 g/mol

mass = 0.0000025152*137.327

mass = 0.00034540487 g of Ba+2

if we need mg so, multiply by 1000

mass = 0.00034540487*1000 = 0.34540487 mg of Ba+2

B.

Assume that

Zn(CH3COO)2 (s) <--> Zn+2(aq)+ 2CH3COO-(aq)

then, the equilibrium definition is given as:

Ksp = [Zn+2][CH3COO-]^2

recall that solids has activity of 1, therefre they must not be included in euqilibirum expression

Ksp = [Zn+2][CH3COO-]^2

Assume that, due to stoichioemtry

Zn(CH3COO)2 = Zn+2(aq) = 2CH3COO-(aq)

If Zn+2 = S

CH3COO- = 2S

then

Ksp = (S)*(2S)^2

Ksp = (S)(4S^2) = 4*S^3

Ksp = 4*S^3

we need to get S in mol per liter so

43 g/L --> mol/L

change grams of Zinc acetate to mol of Zinc acetate

MW Zn(CH3COO)2 = 183.48 g/mol

then

1 mol = 183.48 g

x mol = 43 g

mol = 43/183.48 = 0.23435796 mol of zinc acetate

since we have this per liter, no need to modify volume

S = 0.23435796 mol/L

now

Ksp = 4*S^3

Ksp = 4*(0.23435796)^3 = 0.05148

Ksp = 5.15*10^-2

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