Determination of the solubility- Product Constant for a Sparingly soluble salt N

Question

Determination of the solubility- Product Constant for a Sparingly soluble salt Need the answer to the post lab questions below

QUESTIONS 1 ithe standard solutions had unknowinglybeen made up to be 0004 M AgNO, and 0000 M K,cro, would this have affected your results? How? 2 If your cuvette had been dirty, how would this have affected the value of K,? 1 Using your determined value of K, calculate how many milligrams of Ag,CrO, will dissolve in 10.0 mL of H,O. The experimental procedure for this experiment has you add 5 mL of 0004 M AgNO, to 5 mL of 0.0024 MKCrO, Is either of these reagents in excess, and if so, which one? 5. Use bae your experimentally determined value of Kp and show, by calculation, that Ag,CrO, should itate precip- when 5 mL of 0.004 M AgNO, are added to 5 mL of 0,0024 M K,CrO,

Explanation / Answer

1) 2AgNO3 + K2CrO4 --> Ag2CrO4(s) + 2KNO3

If the standard solutions had unknowingly been made up to be 0.0024 M AgNO3 and 0.0040 M K2CrO4, by mixing them Ag2CrO4 precipitated out and left 0.0024M KNO3 mixed with

(0.0040 - 0.0024/2)M,

(0.0040 – 0.0012) = 0.0028M, of K2CrO4

2) The more dirt on the cuvette, the higher the absorbance reading, so would have a higher apparent concentration. The fact that light cannot pass through a dirty cuvette will alter the Ksp. The Ksp would have been larger.

3.

Molar mass of Ag2CrO4 = 331.73 g/mol

Ksp of Ag2CrO4 = 2.88 x 10^-12

10.0 mL of H2O = 0.0100L of H2O

The Ksp equation:

Ksp = [Ag+]2[CrO42-]

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

x = 8.96 x 10^-5 M

g of Ag2CrO4 will dissolve in 10.0 mL of H2O = 8.96 x 10^-5 mol/L x 0.0100 L x 331.73 g/mol = 0.000297 g

mg of Ag2CrO4 will dissolve in 10.0 mL of H2O = 0.297 mg

4)

m moles of K2CrO4 = 5mL x 0.0024 mol/1000 mL = 0.012mmol

m moles of AgNO3 = 5mL x 0.0040 mol/1000 mL = 0.020mmol

AgNO3 should be twice as many moles of K2CrO4 = 0.012mmol x 2 = 0.024 mmol

But there is only 0.020mmol AgNO3, so it is limited. The reagents in excess is K2CrO4

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.