InClass02 Ksp 105F96

Ag₂CrO₄ = 2Ag⁺ + CrO₄^2-

K_sp = [Ag⁺]²[CrO₄^2-] = 1.2 x 10^-12

B. Calculate the concentrations of Ag⁺ and of CrO₄^2-.

Since there is no source of silver or of chromate outside of the dissociation of silver chromate itself, this is a simple solubility problem.

From Charge Balance [Ag⁺] = 2[CrO₄^2-]. Let s = [CrO₄^2-] ; then [Ag⁺] = 2s . Plugging into the K_sp expression, (2s)²s = K_sp, or 4s³ = K_sp, or s = (K_sp/4)^1/3. Putting in the values:

You may need to find out how to take the cube root on your calculator. Look up how to use the y^x button.

[Ag⁺] = 2s = 2(6.69 x 10^-5) = 1.34 x 10^-4.

2. A solution known to have [Ag⁺] = 0.045 M has a pile of silver chromate in the bottom of it. Compute the concentration of chromate [CrO₄^2-] in the solution.

Since the [Ag⁺] has been determined by addition from the outside, probably as AgNO₃, this is a common ion problem. From the K_sp expression in problem 1A above, we have:

Note that if silver nitrate had been the source of the excess silver, and if the problem had stated that C_Ag+ = 0.045 M then charge balance would be:

[Ag⁺] = 2[CrO₄^2-] + [NO₃^-]

[NO₃^-] would be large, 0.045 M, but chromate would be very small, about

5.9x10^-10 M, so [Ag⁺] = [NO₃^-] = 0.045 M to a very good approximation.

3. 100.0 mL of 0.100 M AgNO₃ and 50.0 mL of 0.300 M NaCl are mixed together. Complete the following Initial/Final Spreadsheet, finally calculating the concentrations of all species in solution after reaction has taken place. K_sp AgCl is 1.82x10^-10.

Balanced Reaction: Ag⁺ + NO₃^- + Na⁺ + Cl^- = AgCl(s) + Na⁺ + NO₃^-

	millimoles Ag+	millimoles NO3-	millimoles Na+	millimoles Cl-	millimoles AgCl(s)	Volumes
Initial	10.0 mmoles	10.0 mmoles	15.0 mmoles	15.0 mmoles	none yet	100 + 50 mL
Final	~0	10.0 mmoles	15.0 mmoles	5.0 mmoles	10.0 mmoles precipitate	150 mL
Molar Conc.	5.47x10-9M	0.0667 M	0.100 M	0.0333 M	not in solution

Initial millimoles:

Since AgNO₃ ionizes completely in solution ("all nitrates are soluble"),

mmole Ag⁺ = mmoles NO₃^- = 100 mL x 0.100 mmole AgNO₃/mL = 10.0 mmole Ag⁺ and NO₃^-

Since all sodium salts are soluble, NaCl ionizes completely in solution, so

mmole Na⁺ = mmole Cl^- = 50 mL x 0.300 mmole NaCl/mL = 15.0 mmole Na⁺ and Cl^-

Final millimoles:

Silver is the limiting reagent, so it goes to approximately zero.

Chloride is reduced from 15 mmoles to 5 mmoles after the 1:1 reaction with silver.

10.0 mmoles of AgCl(s) are formed as a precipitate that sits at the bottom of the beaker.

The spectator ions remain unchanged.

Final Molar Concentrations:

C_NO3- = 10.0 mmoles/150 mL = 0.0667 M

C_Na+ = 15.0 mmoles/150 mL = 0.100 M

C_Cl- = 5.0 mmoles/150 mL = 0.0333 M

then

[Ag⁺] = K_sp/[Cl^-]; appx. = K_sp/C_Cl- = 1.82x10^-10/0.0333 = 5.47x10^-9 M (common ion problem)

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