DO NOW: What is dissolution? What is precipitation? How are they related? What does the term saturated mean?

What happens here? Water NaCl AgCl Na+ Cl- Na+ Cl- Na+ Cl- NaCl(s) -> Na + (aq) + Cl - (aq) AgCl(s) Ag + (aq) + Cl - (aq) AgCl Ag+ Cl- Cl- Ag+ Na+ Cl- Na+ Cl- Na+ Cl-

What’s the difference? Water NaCl AgCl Na+ Cl- Na+ Cl- Na+ Cl- NaCl(s) -> Na + (aq) + Cl - (aq) AgCl(s) Ag + (aq) + Cl - (aq) AgCl Ag+ Cl- Cl- Ag+ Na+ Cl- Na+ Cl- Na+ Cl- COMPLETELY SOLUBLE FOR “INSOLUBLE” OR SLIGHTLY SOLUBLE SOLIDS

Dissolution vs. Precipitation

These opposing processes, dissolution and precipitation will continue until … The system reaches solubility equilibrium At this point the rate of precipitation equals the rate of dissolution No net change in concentration results The solution is saturated

AgCl (s) ↔ Ag + (aq) + Cl - (aq) Solubility product constant- Will just be a product of two species. Could be raised to a power depending on coefficients. K = [Ag + ][Cl - ] sp =1.6 x Since this reaction reaches equilibrium, apply equilibrium concepts What does the size of K indicate? Always write solid on left and ions on right.

The bigger the K… The more ions present… The more soluble the solid is. Solutes like NaCl would have such a large K the reaction goes to completion (one way)

1.Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH) 2 are used in industry as a strong, inexpensive base. Calculate the molar solubility of Ca(OH) 2 in water if the K sp is 6.5x When lead(II) fluoride (PbF 2 ) is shaken with pure water at 25°C, the solubility is found to be 0.64 g/L. Calculate the K sp of PbF 2. Work and answers follow. So try first!!!

Problem 1: 1. Write eqn and expression. 2. Set up Ice. 3. Solve Ca(OH) 2 (s) Ca 2+( aq) + 2OH - (aq) I00I00 Cx 2x E x 2x Ksp =[Ca 2+ ][OH - ] 2 = so x = molar solubility = M 4x 3 = 6.5x10 -6

Problem 2: 1. Write rxn and expression. 2. Solubility means at eqn, so 0.64 g/L is the equilibrium qty 0.64 g/L convert to M PbF 2 (s) Pb 2+( aq) + 2F - (aq) I00I00 Cx 2x E x =.00260M 2x = 2( M) Ksp =[Pb 2+ ][F - ] 2 = 7x10 -8

A common laboratory method for preparing a precipitate is to mix solutions containing the component ions. Does a precipitate form when L of 0.30 M Ca(NO 3 ) 2 is mixed with L of M NaF? First, what will potentially ppt? Write a eqn and see which product could be insoluble or slightly soluble. Then consider the equilibria for that solid. You will need to look up a Ksp on p. 718 See what you can do… work and answer follows. Try first!!!

Ca(NO 3 ) 2(aq) +2NaF (aq) ->CaF 2(?) +2NaNO 3(?) We know all alkali metals and nitrates are completely soluble, so only choice is CaF2. Will CaF2 ppt? Let’s write a reaction. CaF 2(s) Ca 2+ (aq) + 2F - (aq) Now “will a ppt form” really means will rxn go to left…sounds like a Q problem to me. Will Q>Ksp? Work it out… remember the dilution… Ksp for CaF 2 = 4.0x [Ca 2+ ]=(.100 L)(0.30M)/(.100L +.200L) = 0.100M [F - ]=(.200 L)(0.060M)/(.100L +.200L) = 0.040M Q = [Ca 2+ ][F - ] 2 = 1.6x So Q>>K, meaning conc. of ions is too high so rxn will proceed to the left to use ions, thereby precipitating, so YES!

Name, FormulaK sp Aluminum hydroxide, Al(OH) 3 Cobalt(II) carbonate, CoCO 3 Iron(II) hydroxide, Fe(OH) 2 Lead(II) fluoride, PbF 2 Lead(II) sulfate, PbSO 4 Silver sulfide, Ag 2 S Zinc iodate, Zn(IO 3 ) 2 3x x x x x x x Mercury(I) iodide, Hg 2 I 2 3.9x Which is more soluble, Fe(OH) 2, PbF 2, or Ag 2 S? 2. Which is more soluble, PbF 2, or PbSO 4 ? The higher the K sp value, the greater the solubility, as long as we compare compounds that have the same total number of ions in their formulas. All three of these solutes break up exactly the same way, namely AB 2 A B - or A 2 B 2A + + B 2- meaning Ksp = 4x 3 where x is solubility. So you can simply look at their Ksp and whichever has larger Ksp, has larger x ( greater conc at eq), therefore most soluble. So Ag2S<Fe(OH)2< PbF2 This comparison is different. PbF 2 Pb F - so Ksp = 4x 3 = 3.6x10 -8 PbSO 4 Pb 2+ + SO 4 2- meaning Ksp = x 2 = 1.6x10 -8 So solve for x or estimate… cube root has to be larger x than sq root, so PbF2 again.

How can we get more calcium hydroxide to precipitate ? Common ion effect - Presence of common ion shifts equilibrium to left to consume ion ( ppt occurs) Many choices of “stresses” here, but we need a source of calcium or hydroxide ions The common ion lowers the solubility by of a slightly soluble substance What is its solubility in 0.10 M Ca(NO 3 ) 2 ? K sp of Ca(OH) 2 is 6.5x Try this! Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq)

AS BEFORE 1. Write eqn and expression. 2. Set up ICE but this time you have an init. conc. 3. Solve Ca(OH) 2 (s) Ca 2+( aq) + 2OH - (aq) I0.10M 0 Cx 2x E 0.1+x 2x Ksp =[Ca 2+ ][OH - ] 2 = (0.1 + x)(2x) 2 = 6.5x10 -6 Can we ignore this x? (0.1)(2x) 2 = 6.5x10 -6 so x = molar solubility = M ( check assump. to see if within 5% of Ksp) So compared to slide 10, where we calculated the solubIlity of Ca(OH)2 in water to be M … we have decreased solubility here by adding common ion which shifted rxn to the left, thereby ppting.

Specifically, how can the solubility of Mg(OH) 2 be increased? Mg(OH) 2(s) Mg 2+ (aq) + 2OH - (aq) Write an equation for the dissolution of the sparingly soluble calcium carbonate. Will the addition of HCl effect this system? How? You did this in the LeChat lab with milk of magnesia. We added vinegar, a source of H+. You can add any acid as the H+ will consume OH- causing reaction to shift right, thereby Increasing solubility, greater conc of ions. (1) CaCO 3 (s) Ca 2+ (aq) + CO 3 2- (aq) HCl will provide H+. Will any ions above react with H+… do you see any conj bases? (2) H + (aq) + CO 3 2- (aq) -> HCO 3 - (aq) So H+ will react with and thus reduce the conc of carbonate ions as seen in rxn 2. As carbonate ions were removed (in rxn 1) that reaction will shift right producing more ions, increasing the solubility… significance? Calcium carbonate is more soluble in acid than in water.

Selective Precipitation Selective precipitation is used to separate a solution containing a mixture of ions. A precipitating ion is added to the solution until the Q sp of the more soluble compound is almost equal to its K sp. The less soluble compound will precipitate in as large a quantity as possible, leaving behind the ion of the more soluble compound. A solution consists of 0.20 M MgCl 2 and 0.10 M CuCl 2. Calculate the [OH - ] that would separate the metal ions as their hydroxides. K sp of Mg(OH) 2 = is 6.3x ; K sp of Cu(OH) 2 is 2.2x

1.Ok. So we are trying to separate these metallic ions, Mg 2+ and Cu 2+. Look at their Ksp’s. Which solid would ppt first if OH- were added to the container? Consider which solid is less soluble? 2.Since Ksp of Cu(OH)2 is so much smaller it is less soluble, meaning if OH- is added it will ppt with Cu2+ much more readily than with Mg2+ (which is more soluble) 3.So let’s see what the max [OH-] we could add without causing Mg2+ to ppt. 4.Ppt starts when Q Mg(OH)2 > 6.3x [Mg 2+ ][OH - ] 2 = (.20)x 2 >6.3x [OH-] cannot exceed 5.6x10 -5 M or Mg(OH) 2 will also start ppting. 6.Now, let’s see if the separation was effective. It is if 99% separation occurred. 7.If we did add this conc of [OH-], let’s see how much [Cu2+] would still be left in solution. [Cu 2+ ][OH - ] 2 = x(5.6x10 -5 ) 2 =2.2x [Cu2+]= 7 x M So looks good because that is <<< than 1% of the original 0.1 M Cu2+. So more than 99% was separated as ppt. An alternate way to proceed after step 5 is as follows… 6b. Let’s see what [OH-] needs to be to ppt the Cu(OH)2. 7b. [Cu2+][OH-] 2 = (.10)x 2 >2.2x b. A conc in excess of [OH-] = 4.7 x M will cause ppt of Cu 2+ only as long as we don’t approach [OH-]= 5.6 x M Mg 2+ Cl - Cu 2+ Mg 2+ Cu 2+ Source of OH-

Ok you made it. Now try the problems!