Solubility Homework Set Worked Out Solutions

# Whatisthemolarsolubilityforpbcro4ksppbcro428x1013

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Unformatted text preview: ange” rows (i.e. the sum): _____Cr3+(aq) + 3F‐(aq) CrF3(s) Initial ‐‐‐‐‐ 0 M 0 M Change ‐‐‐‐‐ +1x M +3x M Equilibrium ‐‐‐‐‐ +x M +3x M These equilibrium concentrations can be substituted into the solubility product expression. Since Ksp is given, we can solve for x, which is the molar solubility: Ksp=[Cr3+] [F‐]3= (x)(3x)3= 27x4= 6.6 x 10‐11 x= 1.3 x 10‐3 M 6. What is the molar solubility for PbCrO4? Ksp PbCrO4 = 2.8 x 10‐13 Initial Change Equilibrium PbCrO4(s) ____Pb2+(aq) + CrO42‐(aq) ‐‐‐‐‐ 0 M 0 M ‐‐‐‐‐ +1x M +1x M ‐‐‐‐‐ +x M +x M Ksp=[Pb2+] [CrO42‐]= (x)(x)= x2= 2.8 x 10‐13 x= 5.3 x 10‐7 M 7. The solubility of BaF2 is 1.3 g/L. What is the solubility product constant? Conversely, Ksp can be calculated from either solubility or molar solubility. First, convert solubility in g/L to mol/L (M) since the concentrations used in Ksp are in mol/L (M). If the given solubility is already in mol/L, this step is unnecessary: 1.3 g BaF2 _ x 1 mol BaF2__ = 0.0074 mol/L= 0.0074 M BaF2 1 L 175.33 g BaF2 From this molar concentration and the equilibrium expression, we can calculate the equilibrium concentrations of each ion using their respective coefficients in the equilibrium: Ba2+(aq) + 2F‐(aq) BaF2(s) 0.0074 M BaF2 x 1 mol Ba2+ = 0.0074 M Ba2+ 1 mol BaF2 2 Dr. Fus CHEM 123 0.0074 M BaF2 x 2 mol F‐ = 0.0148 M F‐ 1 mol BaF2 With these concentrations and the solubility‐product expression, we can calculate Ksp. Don’t forget the exponents! Ksp=[Ba2+] [F‐]2=(0.0074)(0.0148)2=1.6 x 10‐6 8. One liter of a saturated solution of silver sulfate contains 4.5 g of Ag2SO4. Calculate the solubility product constant for Ag2SO4. 4.5 g Ag2SO4 x 1 mol Ag2SO4__ = 0.0144 mol/L= 0.0144 M Ag2SO4 1 L 311.81 g Ag2SO4 2Ag+(aq) + SO42‐(aq) Ag2SO4 (s) 0.0144 M Ag2SO4 x 2 mol Ag+ = 0.0288 M Ag+ 1 mol Ag2SO4 0.0144 M Ag2SO4 x 1 mol SO42‐ = 0.0144 M SO42‐ 1 mol Ag2SO4 +]2 [SO 2‐]=(0.0288)2(0.0144)=1.2 x 10‐5 Ksp=[Ag 4 9. The solubility of copper(II) iodate, Cu(IO3)2, is 1.3 g/L at 25 oC. Calculate the solubility product constant for copper(II) iodate. Ksp=1.2 x 10‐7 10. Calculate the Ksp value for Bi2S3, which has a solubility of 1.0 x 10­15 mol/L at 25oC. Ksp=1.1 x 10‐73 11. The solubilty of lead (II) chloride is 1.6 10­2 M. What is the Ksp of PbCl2? Ksp=1.6 x 10‐5 12. The solubility of CaCO3 (limestone) is 9.5 mg in 1800 mL. What is the Ksp of CaCO3 (Formula Weight = 100.1 g/mol)? If mg or mL is given instead of g or L, an extra step is required to ensure correct units are used: 9.5 mg x 1 g x 1000mL = 0.00528 g/L… 1800 mL 1000 mg 1 L Ksp=2.8 x 10‐9 13. The Ksp for CaF2 is 3.9 x 10‐11. What is the solubility of CaF2 in water in grams/liter? This problem is solved just like problems 5 and 6. However, molar solubility (M=mol/L) must be converted into g/L using the molecular weight of the solid: …2.14 x 10‐4 mol CaF2 x 78.08 g CaF2 = 1.6 x 10‐2 g/L 14. The Ksp for Sn(OH)2 is 2.0 x 10‐26. What is the...
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## This homework help was uploaded on 03/28/2014 for the course CHEM 123 taught by Professor Larosa during the Spring '06 term at Ohio State.

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