where there are only free ions this free ion concentration is referred to as

Where there are only free ions this free ion

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where there are only free ions, this free ion concentration is referred to as activity . Ksp is like the ideal gas law, which is only for gases not attracted to other gases. Ksp works for “ideal” solutes only. In a sense, there are two concentrations of ions, the stoichiometric concentration and the activity. K sp is based on activity, the effective concentrations of ions regarding solubility.)
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Chem 162-2016 Chapter 18 Acid-base & solubility equilibria lecture notes 41 IMPORTANCE OF SOLUBILITY IN INDUSTRY AND MEDICINE BaSO 4 , an insoluble salt, shows pathway blockage of digestive tract. CaC 2 O 4 , calcium oxalate, is 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.
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Chem 162-2016 Chapter 18 Acid-base & solubility equilibria lecture notes 42 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 ) a = (1.2 x 10 52 ) 2 b = (3.6 x 10 15 ) K = (a)(b) log(K) = 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) = 104.1580 + 15.5563 = 119.7143 log(K) = 119.7143 log(K) = 119 + 0.7143 log(a) = b + c a = 10 b x 10 c K = 10 0.7143 x 10 119 K = 5.18 x 10 119
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Chem 162-2016 Chapter 18 Acid-base & solubility equilibria lecture notes 43 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. (a) FePO 4 (s) Fe 3+ (aq) + PO 4 3- (aq) [Fe 3+ ][PO 4 3- ] = K sp (b) Cr(OH) 3 (s) Cr 3+ (aq) + 3OH - (aq) [Cr 3+ ][OH - ] 3 = K sp *The product of the equilibrium concentration of the ions in the solution is equal to K, the equilibrium constant (= the solubility product constant = K sp ).
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Chem 162-2016 Chapter 18 Acid-base & solubility equilibria lecture notes 44 SOLUBILITY EQUILIBRIA p.5.CHEM162-2010 8th WEEK RECITATION ET Note: Given the solubility of a salt, find K sp Z&Z 75 Use the following data to calculate the K sp value for each solid.
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