CaC 2 O 4 calcium oxalate are main component of kidney stones Ca2 is in the

# Cac 2 o 4 calcium oxalate are main component of

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CaC 2 O 4 , calcium oxalate, are main component of kidney stones. Ca 2+ is in the blood, which reacts with oxalates in certain fruits and vegetables, e.g., rhubarb and spinach. Treatment: adjust diet. Ca 5 (PO 4 ) 3 OH, hydroxyapatite, is tooth enamel. Ca 5 (PO 4 ) 3 OH + H 3 O + Ca 5 (PO 4 ) 3 + + 2H 2 O Tooth enamel is soluble in acid. Ca 5 (PO 4 ) 3 OH + F - Ca 5 (PO 4 ) 3 F (fluoroapatite) + OH - Ca 5 (PO 4 ) 3 F + H 3 O + Ca 5 (PO 4 ) 3 + + HF + H 2 O Fluoroapatite less soluble in acid.
CHEM162-2013 CHAPTER 17 37 HOW TO WORK WITH LARGE NUMBERS Some calculations are too big (or too small) for your calculator; i.e., when K is very large or very small, your calculator may not be able to handle it. So work with logs. e.g.: K = (1.2 x 10 52 ) 2 (3.6x10 15 ) log(ab) = log(a) + log(b) log(K) = log((1.2x10 52 ) 2 ) + log(3.6x10 15 ) log(K) = 2 x log(1.2x10 52 ) + log(3.6x10 15 ) log(K) = 119.7147 log(K) = 119 + 0.7147 log(a) + log(b) = log(ab) (???) K = 10 0.7147 x 10 119 K = 5.18 x 10 119
38 Solubility Product Constants AgCl(s) Ag + (aq) + Cl - (aq) This is the equilibrium between a saturated solution of AgCl and solid AgCl. All of the AgCl that dissolves is dissociated into Ag + and Cl - ions. Equilibrium expression: K sp = [Ag + ][Cl - ] Pure liquids and pure solids don’t appear in equilibrium constant expressions. K sp , the solubility product constant, is the product of the concentrations of the ions involved in the solubility equilibrium, each raised to a power equal to the stoichiometeric coefficient of that ion in the chemical equation for the equilibrium. A(s) 3B(aq) + 2C(aq) A(s) B(aq) + B(aq) + B(aq) + C(aq) + C(aq) ([B] 3 [C] 2 ) = K sp H&P Example 16.1 mod.: Write the equilibrium equations and equilibrium constant expressions for (a) iron(III) phosphate and (b) chromium(III) hydroxide.
39 SOLUBILITY EQUILIBRIA p.5.CHEM162-2010 8th WEEK RECITATION ET Note: Given the solubility of a salt, find KspZ&Z 75 Use the following data to calculate the Kspvalue for each solid. a. The solubility* of CaC2O4is 4.8 x 10-5mol/L. CaC2O4(s) Ca2+(aq) + C2O42-(aq) CaC2O4(s)Ca2+(aq) + C2O42-(aq) Initial Y 0 Change -X +X Equilibrium Y - X +X CaC2O4(s)Ca2+(aq) + C2O42-(aq) Initial Y 0 Change -4.8 x 10-5+4.8 x 10-5+4.8 x 10-5 0 +X X 0 b. The solubility of BiI3is 1.32 x 10-5mol/L. BiI 0 +3X 3X 0 ) )
CHEM162-2013 CHAPTER 17 40 From Kimmel’s notes ET Note: Given the solubility of a salt, find Ksp; the solubility of the salt is based on its pH. A saturated calcium hydroxide solution has a pH of 12.14. Calculate the K sp of calcium hydroxide. pH = 12.14 pOH = 14.00 - 12.14 = 1.86 [OH - ] = 10 -pOH = 10 -1.86 = 1.38 x 10 -2 Ca(OH) 2 (solid) Ca 2+ (aq) + 2OH - (aq) Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) Initial Y 0 0 Change Equilibrium 1.38 x 10 -2 Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) Initial Y 0 0 Change -X +X +2X Equilibrium Y-X +X 1.38 x 10 -2 0 + 2X = 1.38 x 10 -2 X = 0.0069 Ca(OH) 2 (s) Ca

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