Common Ion Effects. CH 3 CO 2 (aq) + Na + (aq)
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1 Common Ion Effects If two reactions both involve the same ion, then one reaction can effect the equilibrium position of the other reaction. The ion that appears in both reactions is the common ion. Buffers are made by combining an acid with its conjugate base. CH 3 CO 2 H(aq) CH 3 CO 2 (aq) + H + (aq) CH 3 CO 2 Na(aq) CH 3 CO 2 (aq) + Na + (aq) The acetic acid contains an acetate ion. Sodium acetate also contains an acetate ion. Sodium acetate and acetic acid share a common ion, the acetate ion, so the addition of sodium acetate can affect an acetic acid equilibrium.
2 Common Ion Effects If two reactions both involve the same ion, then one reaction can effect the equilibrium position of the other reaction. The ion that appears in both reactions is the common ion. Buffers are made by combining an acid with its conjugate base. CH 3 CO 2 H(aq) CH 3 CO 2 (aq) + H + (aq) CH 3 CO 2 Na(aq) CH 3 CO 2 (aq) + Na + (aq) The acetic acid contains an acetate ion. Sodium acetate also contains an acetate ion. Sodium acetate and acetic acid share a common ion, the acetate ion, so the addition of sodium acetate can affect an acetic acid equilibrium. Any common ion Well, the same thing happens in all equilibria. For CuBr K sp = 4.2 x 10-8 The solubility of CuBr will be lower in a 0.05 M NaBr solution than it would be in pure water.
3 Let s start with a saturated CuBr solution. In a saturated CuBr solution 4.2 x 10-8 = [Cu + ][Br ] [Cu + ] = [Br ] = M Now, what is the solubility of CuBr in a M NaBr solution?
![2 x 10-8 = [Cu + ][Br ] [Cu + ] = [Br ] = 0.](/docs-images/46/21535020/images/page_3.jpg "000205 M Now, what is the solubility of CuBr in")
4 Let s start with a saturated CuBr solution. In a saturated CuBr solution 4.2 x 10-8 = [Cu + ][Br ] [Cu + ] = [Br ] = M Now, what is the solubility of CuBr in a M NaBr solution? Simply set up an equilibrium table that has a starting conc of Br = to M... CuBr Cu + Br start solid change + x + x equilibrium x x 4.2 x 10-8 = (x)( x) small x approx 4.2 x 10-8 = (x)(0.050) x = 8.4 x 10-7 [Cu + ] = 8.4 x 10-7 M
5 Determine the solubility of CaSO 4, K sp = 6.1 x 10-5, in a M Na 2 SO 4 solution. Determine the solubility of CaCl 2 in a M Na 2 SO 4 solution.
6 Determine the solubility of CaSO 4, K sp = 6.1 x 10-5, in a M Na 2 SO 4 solution. CaSO 4 (s) Ca 2+ (aq) + SO 4 2- (aq) solid solid +x +x solid x x x ( x) = 6.1 x 10-5 Solubility is M Determine the solubility of CaCl 2 in a M Na 2 SO 4 solution.
7 Determine the solubility of CaSO 4, K sp = 6.1 x 10-5, in a M Na 2 SO 4 solution. CaSO 4 (s) Ca 2+ (aq) + SO 4 2- (aq) solid solid +x +x solid x x x ( x) = 6.1 x 10-5 Solubility is M Determine the solubility of CaCl 2 in a M Na 2 SO 4 solution. (K sp of CaCl 2 >> K sp CaSO 4 ) CaSO 4 (s) Ca 2+ (aq) + SO 4 2- (aq) solid solid +x 0 solid x x (0.0050) = 6.1 x 10-5 x = Solubility is M
8 Precipitation What concentration of NaCl is required to precipitate AgCl from a 0.10 M AgNO 3 solution?
9 Precipitation What concentration of NaCl is required to precipitate AgCl from a 0.10 M AgNO 3 solution? NaCl(s) H 2 O Na + (aq) + Cl (aq) K > 10 AgNO 3 (s) H 2 O Ag + (aq) + NO 3 (aq) K > 10 So, all of the NaCl and the AgNO 3 dissolves and dissociates. NaCl(aq) Na + (aq) + Cl (aq) AgNO 3 (aq) Ag + (aq) + NO 3 (aq)
10 Precipitation What concentration of NaCl is required to precipitate AgCl from a 0.10 M AgNO 3 solution? NaCl(s) H 2 O Na + (aq) + Cl (aq) K > 10 AgNO 3 (s) H 2 O Ag + (aq) + NO 3 (aq) K > 10 Since we are mixing two solution, we know all of the NaCl and the AgNO 3 is dissolved and dissociated. NaCl(aq) Na + (aq) + Cl (aq) AgNO 3 (aq) Ag + (aq) + NO 3 (aq) Now, we have to worry about the possibility of a metathesis reaction (ions switching) occurring. H 2 O NaNO 3 (s) Na + (aq) + NO 3 (aq) K > 10 AgCl(s) H 2 O Ag + (aq) + Cl (aq) K = 1.6 x Yes, AgCl can precipitate because it is not very soluble
11 The reaction is AgCl(s) Ag + (aq) + Cl - (aq) K sp = 1.6 x solid 0.10? and if Q K then a precipitate will form.
12 The reaction is AgCl(s) Ag + (aq) + Cl - (aq) K sp = 1.6 x solid 0.10? and if Q K then a precipitate will form. [Ag + ] o [Cl ] o 1.6 x [Cl ] o 1.6 x [Cl ] o 1.6 x 10-9 So, if [NaCl] 1.6 x 10-9 then Q K and a precipitate will form. (There is enough Cl - in tap water to cause AgCl to precipitate from a AgNO 3 solution.)
![6 x 10-10 0.10 [Cl ] o 1.6 x 10-10 [Cl ] o 1.6 x 10-9 So, if [NaCl] 1.](/docs-images/46/21535020/images/page_12.jpg "6 x 10-9 then Q K and a precipitate will form.")
13 Common ion and ph dependence The solubility of some materials displays a ph dependence This is a common ion effect. for example... For Cu(OH) 2 K sp = 2.2 x Cu(OH) 2 (s) Cu 2+ (aq) + 2 OH (s)
14 What is the solubility of Cu(OH) 2 in neutral water? K sp = [Cu 2+ ][OH ] 2 = 2.2 x Set up a table, before anything dissolves [Cu 2+ ] = 0, and [OH ] ~ 0 right? Cu(OH) 2 Cu 2+ OH solid 0 ~0
15 What is the solubility of Cu(OH) 2 in neutral water? K sp = [Cu 2+ ][OH ] 2 = 2.2 x Set up a table, before anything dissolves [Cu 2+ ] = 0, and [OH ] ~ 0 right? Cu(OH) 2 Cu 2+ OH solid 0 ~0 NO, OH is very small, but it is not zero. In neutral water [OH ] = 10-7 M. This reaction goes only slightly toward the products, so a concentration of 10-7 for one of the products is going to be significant! Cu(OH) 2 Cu 2+ OH initial solid change + x + 2 x equilibrium x x
![Cu(OH) 2 Cu 2+ OH solid 0 ~0 NO, OH is very small, but it is not zero. In neutral water [OH ] = 10-7 M.](/docs-images/46/21535020/images/page_15.jpg "This reaction goes only slightly toward the products, so a concentration of 10-7 for one of the products is going to be")
16 What is the solubility of Cu(OH) 2 in neutral water? K sp = [Cu 2+ ][OH ] 2 = 2.2 x Set up a table, before anything dissolves [Cu 2+ ] = 0, and [OH ] ~ 0 right? Cu(OH) 2 Cu 2+ OH solid 0 ~0 NO, OH is very small, but it is not zero. In neutral water [OH ] = 10-7 M. This reaction goes only slightly toward the products, so a concentration of 10-7 for one of the products is going to be significant! Cu(OH) 2 Cu 2+ OH initial solid change + x + 2 x equilibrium x x 2.2 x = x( x) 2 Small X approx? Well, since 10-7 is much bigger than by a factor of about the change, x, is going to be small as compared to 10-7, so let s give the small x approx a try x = x 2.22 x 10-8 = x So, I solved it iteratively, Approximation not valid x 10-8 /10-7 x 100 = 44% x = 2.2 x /( x) 2
17 I plugged in 2.2 x 10-8 for the x on the right, evaluated the expression, and repeated until the the x s converged in 12 steps to 1.36 x e e e e e e e e e e e e e e e e e e e e e e e e e e-08 So, the molar solubility of Cu(OH) 2 = 1.36 x 10 8 M in neutral water.
18 What is the solubility of Cu(OH) 2 if the ph of the solution starts at 12.00? (Remember, if you are given the ph of a solution then you know [H + ], unless is the question says something like starting ph.) Cu(OH) 2 (s) Cu 2+ (aq) + 2 OH (s)
19 What is the solubility of Cu(OH) 2 if the ph of the solution starts at 12.00? (Remember, if you are given the ph of a solution then you know [H + ], unless is the question says something like starting ph) Cu(OH) 2 (s) Cu 2+ (aq) + 2 OH (s) Cu(OH) 2 Cu 2+ OH initial solid 0? change equilibrium
![], unless is the question says something like starting ph) Cu(OH) 2 (s)](/docs-images/46/21535020/images/page_19.jpg "Cu 2+ (aq) + 2 OH (s) Cu(OH) 2 Cu 2+ OH initial solid 0?")
20 What is the solubility of Cu(OH) 2 if the ph of the solution starts at 12.00? (Remember if you are given the ph of a solution then you know [H + ], unless is the question says something like starting ph) Cu(OH) 2 (s) Cu 2+ (aq) + 2 OH (s) Cu(OH) 2 Cu 2+ OH initial solid change + x + 2 x equilibrium x x = [10-12 ][OH ] [OH ] = 10-2
21 What is the solubility of Cu(OH) 2 if the ph of the solution starts at 12.00? (Remember if you are given the ph of a solution then you know [H + ], unless is the question says something like starting ph) Cu(OH) 2 (s) Cu 2+ (aq) + 2 OH (s) Cu(OH) 2 Cu 2+ OH initial solid change + x + 2 x equilibrium x x The change, x, is going to be small as compared to 10-2, so 2.2 x = [Cu 2+ ](10-2 ) 2 [Cu 2+ ] = 2.2 x /10-4 [Cu 2+ ] = 2.2 x Yep, small x approximation is valid. 4.4 x /10-2 x 100 = 4.4 x % The molar solubility went down a great deal when the ph went from 7 to 12!
22 You must be careful while doing these calculations. While the solubility is ph dependent ph is not the only factor. Take the following example... What is the solubility of Cu(OH) 2 in a 0.1 M HNO 3 solution? Find [OH ] K w = [H + ][OH ] = 0.1 [OH ] [OH] = so x = x(10-13 ) x = x 2.2 x 10 4 = x Small x approximation not valid, but lets ignore that fact for a moment and focus and the answer.
23 You must be careful while doing these calculations. While the solubility is ph dependent ph is not the only factor. Take the following example... What is the solubility of of Cu(OH) 2 in a 0.1 M HNO 3 solution? NO Way is the molar solubility 2.2 x OK, is 22,000 not big enough for you? Find [OH ] K w = [H + ][OH ] = 0.1 [OH ] [OH] = so x = x(10-13 ) x = x 2.2 x 10 4 = x That is 22,000 moles of Cu(OH) 2 dissolved in 1 L. This solution would have approximately 2,100,000 g of Cu(OH) 2 in a L of water
24 The problem changed, Cu(OH) 2 (s) + 2 H + (aq) + 2 NO 3 (aq) Cu 2+ (aq) + 2 NO 3 (aq) + 2 H 2 O(l) The HNO 3 reacts with the Cu(OH) 2 to make water and Cu(NO 3 ) 2, and Cu(NO 3 ) 2 is soluble in water Cu(OH) 2 continues to dissolve until the HNO 3 is used up. Once the HNO 3 is used up the [Cu 2+ ] = M The Cu 2+ concentration is determined using the stoichiometry of the neutralization reaction. 0.1 M HNO 3 x 1 mol Cu(NO 3 ) 2 = M Cu(NO 3 ) 2 2 mol HNO 3 Since [Cu 2+ ] = M no more Cu(OH) 2 will dissolve! This was an acid base neutralization in disguise.
25 Less obvious examples of ph dependent solubility The solubility of any salt that dissociates and forms a weak acid or a weak base will demonstrate ph dependence. For example Ag 2 SO 4 is only very slightly soluble in water, K sp for Ag 2 SO 4 = 1.4 x Ag 2 SO 4 (s) 2 Ag + (aq) + SO 4 2 (aq) The solubility of Ag 2 SO 4 will be higher in 1 M HNO 3. Why?
26 Less obvious examples of ph dependent solubility The solubility of any salt that dissociates and forms a weak acid or a weak base will demonstrate ph dependence. For example Ag 2 SO 4 is only very slightly soluble in water, K sp for Ag 2 SO 4 = 1.4 x Ag 2 SO 4 (s) 2 Ag + (aq) + SO 4 2 (aq) The solubility of Ag 2 SO 4 will be higher in 1 M HNO 3. Why? HNO 3 (s) H + (aq) + NO 3 (aq) Since SO 2 2 is a weak base... H + (aq) + SO 4 2 (aq) HSO 4 (aq) The K for this reaction is 1/K a = 1/(1.2x10-2 ) = 83 Ag 2 SO 4 (s) 2 Ag + (aq) + SO 4 2 (aq) So, the presence of H + lowers the concentration of SO 4 2, and, according to Le Chätelier s principle, more Ag 2 SO 4 will dissolve to replace the SO 4 2 that was converted to HSO 4.
27 Common ion and ph dependence The solubility of some materials displays a ph dependence This is a common ion effect. for example... For Cu(OH) 2 K sp = 2.2 x Cu(OH) 2 (s) Cu 2+ (aq) + 2 OH (s) The solubility of any salt that produces an acid or base when it dissolves in water will be ph dependent. NaNO 2 K 2 SO 4 NaCl LiNO 3 The solubility of any metal that can form an insoluble metal hydroxide will be ph dependent. NaCl FeCl 3 Ba(NO 3 ) 2 Cu(NO 3 ) 2
28 Common ion and ph dependence The solubility of some materials displays a ph dependence This is a common ion effect. for example... For Cu(OH) 2 K sp = 2.2 x Cu(OH) 2 (s) Cu 2+ (aq) + 2 OH (s) The solubility of any salt that produces an acid or base when it dissolves in water will be ph dependent. NaNO 2 K 2 SO 4 NaCl LiNO 3 yes yes no no The solubility of any metal that can form an insoluble metal hydroxide will be ph dependent. NaCl FeCl 3 Ba(NO 3 ) 2 Cu(NO 3 ) 2 no yes yes yes
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