A solution was prepared by dissolving 476 mg of silver chromate (Ag2CrO4, MW = 331.73 g mol-1) in 633 mL of water a) moles of Ag2CrO4 b) millimoles of Ag2CrO4 c) molarity of Ag2CrO4 d) molarity of Ag + e) molarity of CrO4 2 - f) parts per million of Ag2CrO4 g) % (weight/volume) Ag2CrO4 (in units of g/mL)

Question

Chemistry

Kason Jarvis

17 December, 09:02

A solution was prepared by dissolving 476 mg of silver chromate (Ag2CrO4, MW = 331.73 g mol-1) in 633 mL of water a) moles of Ag2CrO4 b) millimoles of Ag2CrO4 c) molarity of Ag2CrO4 d) molarity of Ag + e) molarity of CrO4 2 - f) parts per million of Ag2CrO4 g) % (weight/volume) Ag2CrO4 (in units of g/mL)

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Answers (1)

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Peyton Santiago · Answer 1

Silver chromate, Ag2CrO4, a sparingly soluble compound, has a solubility product, Ksp, of 1.2 x 10-12

What is the molar solubility of Ag2CrO4 in pure water?

Explanation:

Use the ICE approach: Write the Ksp equilibrium:

Ag2CrO4 (s) = 2 Ag + (aq) + CrO4 2 - (aq) Ksp = 1.2 x 10 - 12

I : 1 0 0

C : ("0") + 2x + x

E : 1 2x x

Note: above, we let x = [CrO4 2 - ] and so [Ag+] = 2x)

Ksp = [Ag+] 2 [CrO4 2-]/1 = (2x) 2 (x) = 4x3 = 1.2 x 10-12

Or, x3 = 3.0 x 10 - 13 Take cube root of both sides:

=> x = 6.7 x 10 - 5 M = [ CrO4 2 - ] = > [ Ag + ] = 2x = 1.3 x 10 - 4 M