1. State in words is meant by the solubility product equation for Ba(IO3)2: Ksp=

Question

1. State in words is meant by the solubility product equation for Ba(IO3)2:
Ksp= [Ba^2+] [IO3-]^2

2. If 10 mL 0.10M Ba(NO3)2 is mixed with 10mL 0.10 M KIO3 a precipitate forms. Which ion will still be present at appreciable concentration in the equilibrium mixture if Ksp for barium iodate is very small? Indicate your reasoning. What would that concentration be ?
__________ Moles/L
3. Silver sulfate , Ag2SO4, is moderately soluble , and has a Ksp equal to about 1.2*10^-5

a.) What would be the solubility of silver sulfate in water? ________moles/L

b.) What would be its solubility in 0.10 M Na2SO4? _______Moles/L

c.) The difference is caused by what is called the common ion effect. State in words what the common ion effect predicts.

Explanation / Answer

1.

When Ba(IO3)2 dissoves according to the reaction
Ba(IO3)2 (s)<----> Ba2+ (aq) + 2 IO3- (aq)

we get x mol/L of Ba2+ and 2x mol/L IO3-

2.

Ba(NO3)2 + 2KIO3 --> Ba(IO3)2 + 2KNO3

Barium iodate precipitates and potassium nitrate will remain in the solution.

1 mmol of Ba(NO3)2 is mixed with 1 mmol of KIO3, so 0.5 mmol of Ba(IO3)2 will precipitate and 0.5 mmol of Ba(NO3)2 will be present. 1 mmol of KNO3 will also be formed.

The barium concentration in the solution can be thought to come from barium nitrate only; since the Ksp is very small. The Ba(2+) concentration is 0.5/20=0.025 M

NO3(-) concentration comes from barium nitrate and potassium nitrate, so the highest concentration will be for nitrate.
1 mmol NO3 (-) from Ba(NO3)2 and 1 mmol NO3(-) from KNO3, so 2 mmoles of NO3(-) in 20 mL of solution gives 0.1 M NO3(-) concentration.

3.

let x = mol/L os Ag2SO4 that dissolve

This will give 2 x mol/L of Ag+ and x mol/L of SO42-

Ksp = [Ag+]^2[SO42-] = (2x)^2(x) = 4 x^3

x = molar solubility = 1.4 x 10^-2 M


in this case
[Ag+]= 2x
[SO42-]= x + 0.1

1.2 x 10^-5 =( 2x)^2 ( 0.1+x)
x = 6.1 x 10^-3 M

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