5. Use your experimentally determined Laboratory Euperamener 3T3 when 5 mL of 0.

Question

5. Use your experimentally determined Laboratory Euperamener 3T3 when 5 mL of 0.004 value of K and show. by calculation, Agacro, should precipitate M AgNO, are added to s mL of0.0024 that MK Crow. 6. In the back of your textbook, look up the accepted value of K for Agrcros. Calculate the percentage error in your experimentally determined value for Kp. 7. Although Ag2Croa is insoluble in water, is soluble in dilute HNo, Explain using chemical equations. 8. What is the greatest source of error in this experiment? 9. Reviewing the procedure, indicate where careless work could contribute to the error source you identified in question ration, Inc.

Explanation / Answer

2) Moles of Ag+ = (volume of solution in L)*(concentration of AgNO3 in mol/L) = (5 mL)*(1 L/1000 mL)*(0.0040 mol/L) [1 M = 1 mol/L] = 2*10-5 mole (ans).

Moles of CrO42- = (volume of solution in L)*(concentration of K2CrO4 in mol/L) = (5 mL)*(1 L/1000 mL)*(0.0024 mol/L) = 1.2*10-5 mole (ans).

3) The balanced chemical equation is

2 Ag+ (aq) + CrO42- (aq) -------> AgCrO4 (s)

As per the stoichiometric reaction,

2 moles Ag+ = 1 mole CrO42- = 1 mole Ag2CrO4

Therefore, 2*10-5 Ag+ = (2*10-5 mole Ag+)*(1 mole CrO42-/2 moles Ag+) = 1*10-5 mole CrO42-

1.2*10-5 mole CrO42- = (1.2*10-5 mole CrO42-)*(2 moles Ag+/1 mole CrO42-) = 2.4*10-5 mole Ag+.

Since, we do not have 2.4*10-5 mole Ag+ required to react completely with 1.2*10-5 mole CrO42-, hence, Ag+ is the limiting reactant.

Therefore, CrO42- will be in stoichiometric excess and the amount of excess CrO42- = (1.2*10-5 – 1*10-5) mole = 2*10-6 mole (ans).

4) Moles of Ba2+ = (10 mL)*(1 L/1000 mL)*(1*10-5 mol/L) = 1*10-7 mole.

Moles of CrO42- = (10 mL)*(1 L/1000 mL)*(1*10-3 mol/L) = 1*10-5 mole.

Total volume of solution = (10 + 10) mL = 20 mL = (20 mL)*(1 L/1000 mL) = 0.02 L.

[Ba2+] = (moles of Ba2+)/(volume of solution in L) = (1*10-7 mole)/(0.02 L) = 5*10-6 mol/L

[CrO42-] = (1*10-5 mole)/(0.02 L) = 5*10-4 mol/L

The product quotient is Q = [Ba2+]*[CrO42-] = (5*10-6 mol/L)*(5*10-4 mol/L) = 2.5*10-9 M2

Since Q is greater than Ksp, hence a precipitate will form (ans).

5) The dissociation of BaCO3 can be written as

BaCO3 (s) <======> Ba2+ (aq) + CO32- (aq)

Let S be the solubility of Ba2+ in solution. Due to the 1:1 nature of dissociation, the solubility of CO32- = S.

Therefore,

Ksp = [Ba2+][CO32-] = (S)*(S)

===> 5.1*10-9 = S2

===> S = 7.141*10-5

The solubility of Ba2+ ions in the solution is 7.141*10-5 mol/L. Due to the 1:1 nature of dissociation, the solubility of BaCO3 in solution = 7.141*10-5 mol/L.

We have 1000 mL of the solution. We know that 1000 mL = 1 L.

Therefore, moles of BaCO3 in solution = (7.141*10-5 mol/L)*(1 L) = 7.141*10-5 mole.

Molar mass of BaCO3 = 197.34 g/mol.

Therefore, amount of BaCO3 = (197.34 g/mol)*(7.141*10-5 mole) = 0.01409 g 0.014 g.

The solubility of BaCO3 = 0.014 g/1000 mL (ans).

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

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