A saturated solution of silver chromate (Ag2CrO4) was found to have a CrO4^2- io

Question

A saturated solution of silver chromate (Ag2CrO4) was found to have a CrO4^2- ion I concentration of 3.2 times 10^-4 M at 25 degree C . What is: the Ksp for Ag2CrCO4; and solubility of the salt in mg/L? Calculate the molar solubility of lead chloride (PbCh) at 25degreeC in 0.100 M sodium chloride given that Ksp for lead chloride is 1.8 times 10^-4 at 25degreeC. Calculate the concentration of silver ions in solution if exactly 0.150 mol of silver nitrate added to exactly 1 L of 6 M ammonia solution. Kf for [Ag(NH3)2]^+ is 1.7 times 10^7. Dr.Bajue

Explanation / Answer

7. To get a pH of 7.5 for buffer, we need to use a combination which has pKa close to this value,

H2PO4- <==> HPO4-2 + H+     pKa = 7.2

would be used here.

lets say we have 1 M solution of NaH2PO4 the acid and 1 M solution of Na2HPO4 the base

Total molar concentration of buffer is assumed to be 1 M

pH = pKa + log([base]/[acid])

7.5 = 7.2 + log(x/1-x)

1.9953 - 1.9953 x = x

x = 0.666 M

so we would need, 0.666 M of Na2HPO4 and 0.334 M of NaH2PO4 to prepared this buffer

8. with [CrO4^2-] = 3.2 x 10^-4 M

(a) Ksp = [Ag+]^2[CrO4^2-] = (2 x 3.2 x 10^-4)^2(3.2 x 10^-4) = 1.311 x 10^-10

(b) 1.12 x 10^-12 = (2x)^2(x)

x = 6.5 x 10^-5 M

molar solubility of Ag2CrO4 = 6.5 x 10^-5 x 331.73 x 1000 = 21.562 mg/L

9. Ksp = [Pb2+][Cl-]^2

let x be the molar solubility then,

1.8 x 10^-10 = [Pb2+](0.1)^2

[Pb2+] = 1.8 x 10^-8 M is the molar solubility required

10. moles of NH3 added = 6 mols

[Ag(NH3)2] = 0.15 mol/1 L = 0.15 M

Kf = [Ag(NH3)2]+/[Ag+][NH3]^2

[NH3] = 0.6-0.15 = 0.45 M

thus, [Ag+] = 0.15/1.7 x 10^7 x (0.45)^2 = 4.36 x 10^-8 M

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