GENERAL CHEMISTRY WORKSHOP SOLUBILITY EQUILIBRIA 1. A handbook lists the aqueous

Question

GENERAL CHEMISTRY WORKSHOP SOLUBILITY EQUILIBRIA 1. A handbook lists the aqueous solubility of lithium phosphate at 18°C as 0.034 g Li PO/100 f solution. What is the Kg of Li PO, at 18°C? 2. Calculate the molar solubility of lead(lI) iodide in water at 25°C, given that its K 7.1 x 10-9 3. Three drops of 0.20 M KI are added to 100.0 mL of 0.010 M Pb(NO) Will a precipitate 4. A 50.0 mL sample of 0.0152 M Na SO,(ag) is added to 50.0 mL 0.0125 M Ca(NOs)2 (ag). 5.) Should magnesium hydroxide precipitate from a solution that is 0.010 M magnesium chloride 6,) Calculate the molar solubility of Mg(OH) in 1.00 M NHCI (ag). form? For lead(II) iodide Kp 7.1 x 10 (Assume there are 20 drops in one milliliter.) What percentage of the Ca2 remains unprecipitat and also 0.10 M NH,? For magnesium hydroxide Ksp 1.8 x 10 For NHs, K 1.8 x 10 400. mL of 0.5875 M aqueous silver nitrate are mixed with 500. mL of 0.290 M sodium 7. chromate. a) What is the mass of the precipitate? b) What is the concentration of each of the ions or a imations. For silver ions remaining in solution? Please indicate any assumpt chromate Ksp = 9.0 × 10-12 8. The copper(l) ion forms a complex ion with CN according to the following equation Cu+ + 3CN- Cu(CN)s2- Kr 1.0× 101 a. Calculate the solubility of CuBr (s) (Ksp 1.010 b. Calculate the concentration of Br at equilibrium. in 1.0 L of 1.0 M NaCN c. Calculate the concentration of CN at equilibrium. A mixture of PbSO4 (s) and PbS O3 (s) is shaken with pure water until a saturated solution is formed. Both solids remain in excess. What is the [Pb PbSO,. Kp 1.6x 10 for PbS,Os, Ksp 4.0x 10 9. in the saturated solution? For 10. A 2.50 g sample of Ag SO4(s) is added to a beaker containing 0.150 L of 0.025 M BaC1, a) Write an equation for any reaction that occurs. b) Describe the final contents of the beaker; that is, the masses of any precipitates present and the concentration of all the ions remaining in solution.

Explanation / Answer

Q5) Given; Mg(OH)2(s)<----->Mg2+(aq)+2OH- (aq)

Ksp =[Mg2+][OH-]^2=1.8*10^-11

[MgCl2]=0.01M

MgCl2(aq)--->Mg2+(aq)+2Cl-(aq) [fully dissociated]

[Mg2+]=[MgCl2] =0.01M

[NH3]=0.1M

In solution ,NH3 hydrolyses as follows:

NH3 +H2O <-->NH4+ +OH-

Kb=[NH4+][OH-]/[NH3]=1.8*10^-5

Using ICE table ,[OH-] can be calculated out:

Kb=x^2/(0.1M-x) [x<<0.1M ,as kb is very small.so x ca be ignored]

1.8*10^-5=x^2/0.1M

x=[OH-]=1.342*10^-3M

Now ,[OH-]=1.342*10^-3M

[Mg2+]=0.01M

So, [Mg2+][OH-]^2=(0.01M)*(1.342*10^-3M)^2=1.8*10^-8 >Ksp

or , product of ions > Ksp (precipitation will occur) for Mg(OH)2

Ksp=product of solubility of ions in saturated solution

[NH3] [NH4+] [OH-] initial 0.1M 0 0 change -x +x +x equilibrium 0.1M-x x x

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