please answer all three problems. thanks A precipitate of FeS (Ksp = 6.3 times 1

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please answer all three problems. thanks

A precipitate of FeS (Ksp = 6.3 times 10-18) is formed by mixing 125.00 mL of a 0.32 M solution of Fe(NO3)2 with 100.00 mL of a 0.40 M solution of Na2S. What is the minimum number of moles of solid NaCN that must be added to this solution in order to redissolve all of the FeS(s) as the complex ion [Fe (CN)6]4- (for which the overall Kf = 7.7 times 1036)? You have a set of 250.0-mL solutions of ~1.50 times 10-8 M AgNO3, Ba (NO3)2 and Pb (NO3)2. You have a set of 250.0-mL solutions of ~1.50 times 10-8 M AgNO3, Ba (NO3)2 and Pb (NO3)2. To each of these solutions, you add 1.75 times 10-5 moles of K2CrO4. Given that Ksp (Ag2CrO4) = 1.9 times 10-12, Ksp (BaCrO4) = 2.1 times 10-10, and Ksp(PbCrO4) = 1.8 times 10-14, what would you expect to happen? (Use Q sp idea) When NH3 (aq) is added to Ag+ (aq), a precipitate initially forms. What is its formula? (Remember that NH3 (aq) is basic and write this reaction formula first.)

Explanation / Answer

(4)

Moles of Fe+2 = 0.04

Moles os S2- = 0.04

Moles of Fes solid = 0.04

As Ksp is very low so Most of concentration of FeS IS present as solid

FeS(s) = 0.04

CN- Needed = 6*0.04 = 0.24

Moles of NaCN needed = 0.24 moles

(5)Qsp of Ag2CrO4 = 3*(10^ -8)*7*(10^ -5) = 21*(10^ -13)

Ksp of Ag2CrO4 = 1.9*(10^ -12)

Qsp < Ksp

So it is fully soluble

Qsp of BaCrO4 = 2*(10^ -8)*7*(10^ -5) = 14*(10^ -13)

Ksp of BaCrO4 = 2.1*(10^ -10)

Qsp < Ksp

So it is fully soluble

Qsp of PbCrO4 = 2*(10^ -8)*7*(10^ -5) = 14*(10^ -13)

Ksp of PbCrO4 = 1.8*(10^ -14)

Qsp > Ksp

Precipitation occurs

(6)[Ag(NH3)2]+

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