The solubility of a given substance is dependent upon the identity and concentra

Question

The solubility of a given substance is dependent upon the identity and concentration of all substances present in the solution. Below are two solutions in which Cu(OH), is dissolved: in each case, the solubility is not the same as it is in pure water (a) For each solution state whether the solubility of the Cu(OH) in the solution is greater than Question Format: Long or less than that of CulOH), in pure water and briefly account for the change in solubility (i) 1-molar HC(a?) (ii) 1-molar Cu(NO(a? (b) A saturated solution of lead(I) iodide in pure water has a lead ion concentration of 1.3 x 10- mole per liter at 25°C i) Write the dissociation equation when lead(II) iodide is added to pure water (ii) Write the Ksp expression (???) Calculate the value for the solubility product constant of Pb12 at 25°C (c) Calculate the molar solubility of PbI2 in a 0.10-M Pb(NO)2 solution at 25°C. (d) To 333 mL of a 0.12-M Pb( NOb solution, 667 mL of 0.015 M KI is added. Does a precipitate form? Support your answer with calculations. (e) Lead(ll) iodide is a bright yellow solid. The presence of lead ions in a solution can be determined by adding a solution of potassium iodide. Write the net ionic equation for the reaction that results when lead is present.

Explanation / Answer

a) In solution ,Cu(OH)2 dissociates as follows:

Cu(OH)2 (s) <--->Cu2+(aq)+2OH-(aq) [equilibrium equation]

Solubility product constant=Ksp=[Cu2+][OH-]^2

1) in 1M HCl

HCl--->H+ +Cl-

[H+]=1M ,so OH- will be neutralized according to the reaction :

H+ +OH- --->H2O

Thus , a decrease in the concentration of OH- causes the equilibrium of solubility to shift in forward direction ,generating more OH- to counteract the change in its concentration(Le Chatlier's principle)

More Cu2+ also is generated in the process,therby increasing the solubility of Cu(OH)2

2) 1M Cu(NO3)2

Cu(NO3)(s)---> Cu2+(aq) +2NO3-

Now an increase i [Cu2+] shifts the equilibrium in the reverse direction ,decreasing the solubility of Cu(OH)2

b)given [Pb2+]=1.3*10^-3 M

i)PbI2(s) <---> Pb2+(aq)+2I-(aq)

ii) Ksp(PbI2)=[Pb2+][I-]^2

iii) [I-]=2[Pb2+]=2*(1.3*10^-3 M)=2.6*10^-3 M

Ksp=(1.3*10^-3 M)(2.6*10^-3 M)^2=8.788*10^-9

c)

Pb(NO3)2--->Pb2+ +2NO3-

[Pb2+]=[Pb(NO3)2]=0.10M

Ksp=1.4*10^-8=[Pb2+][I-]^2

or, 1.4*10^-8=(0.10M)[I-]^2

[I-]^2=1.4*10^-9

[I-]=3.742*10^-5 M

molar solubiity of PbI2=3.742*10^-5 M

d) mol of Pb(NO3)3=0.333L*0.12mol/L=0.03996 mol=0.04mol

[Pb2+]=[Pb(NO3)2]=0.04mol/(333+667)ml=0.04mol/1000ml=0.04mol/L

mol of KI added=0.667L*0.015mol/L=0.01mol

[I-]=0.01 mol/1L=0.01M

[Pb2+][I-]^2=(0.04M)(0.01M)^2=0.0004

[Pb2+][I-]^2> Ksp(in soluble state) ,so the excess of ions will precipitate as PbI2

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