Question

1. If the standard solutions had unknowingly been made up tobe 0.0024 M AgNO3 and 0.0040 M K2CrO4, would this have affected your results? Why?

2. If your curvette had been dirty, how would this have affected the value of ksp?

3. Using your determined value of ksp, calculate how manymilligrams of Ag2CrO4 will desolve in 10.0 mL of H2O?

4. The experimental procedure for this experiment has you add 5mL of 0.004 M AgNO3 to 5mL of 0.0024 M K2CrO4. Is either of these reagents in excess, if so which one?

5. Use your experimentally determined value of ksp and show,by calculations, that Ag2CrO4 should precipitate when 5mL of 0.004M AgNO3 are added to 5mL of 0.0024 M K2CrO4.

7. Although Ag2CrO4 is insoluble in water, it is soluble indilute HNO3. Explain using chemical equations.

Any help with these would be greatly appreciated!

Answer 1

1. 2AgNO3 + K2CrO4 ==> Ag2CrO4(s) + 2KNO3

Hence, by mixing .0024M AgNO3 and .004M K2CrO4, you would have Ag2CrO4 precipitated out and left you with 0.0024M KNO3 mixed with (0.004-0.0024/2)M, or 0.0028M, of K2CrO4.

2.the more dirt on the cuvette when you take a reading, the higher the OD. this is because OD determines how much light can pass through a sample. the light that passes through the cuvette hits particles in the sample and cannot get through. therefore, the more concentrated a solution, the higher the OD will be. if you have a very dilute concentration of ions and you put this in a dirty cuvette, the OD reading will be applicable to a high concentration of those ions. the fact that light cannot pass through a dirty cuvette will alter the Ksp in that the reading will tell you that you have more soluble product than there actually is.

3. let x = mol/L of Ag2CrO4 that dissolve :

this will give 2x mol/L Ag+ and x mol/L CrO42-

Ksp = 2.88 x 10^-12 = (2x)^2(x) = 4x^3

x = molar solubility = 8.96 x 10^-5 M

8.96 x 10^-5 mol/L x 0.0100 L x 331.73 g/mol= 0.000297 g => 0.297 mg

4. The balanced equation is

2 AgNO3 + K2CrO4 = Ag2CrO4 + 2 KNO3

moles AgNO3 = 5 x 10^-3 L x 0.004 M=2 x 10^-5
moles K2CrO4 = 5 x 10^-3 L x 0.0024 M=1.2 x 10^-5

the ratio between AgNO3 and K2CrO4 is 2 : 1

moles AgNO3 required = 1.2 x 10^-5 x 2 = 2.4 x 10^-5
we have only 2 x 10^-5 moles of AgNO3 so AgNO3 is the limiting reactant and K2CrO4 is in excess

5.Balanced Reaction:

2 Ag+ (aq) + CrO4^2- (aq) --> Ag2CrO4 (s)

Ksp = [Ag+] ^2 * [CrO4-]

If [CrO4-] = x, then [Ag+] = 2x (from the balanced equation)
then Ksp = 2x^2 * x = 4x^3 (substitute in for Ksp equation)

3.83 * 10^-11 = 4x^3
x= 2.12*10^-12

Max Concentration of CrO4^2- that is saturated in solution is: 2.12*10^-12
Max Concentration of Ag^2+ that is saturated in solution is: 2*(2.12*10^-12) = 4.24 * 10^-12

Compare to actual concentration in solution:
Ag2+: 5 mL * (0.004 mol / 1000 mL ) = 2.0 * 10^-5.
CrO4^2-: 5mL * (0.0024 mol / 1000mL) = 1.2 * 10^-5.
Actual amount of both ions in solution is much greater than max amount that will be saturated, so all the excess will precipitate out.

7.silver nitrate & chromic acid are both soluble.

Ag2CrO4 + 2HNO3 = 2AgNO3 + H2O + CrO3