The End of Equilibrium! (well, for us!) TEXT AND YOUR QUESTION TO 37607

K sp What is the solubility of FeCO 3 ? Solubility = MAXIMUM amount of a compound that can dissolve in water. This is actually an equilibrium. TEXT AND YOUR QUESTION TO 37607

Equilibrium problems involve 3 parts: 1. Balanced equation 2. “K-equation” 3. ICE chart What is the balanced equation for dissolving something? TEXT AND YOUR QUESTION TO 37607

FeCO 3 (s)  Fe 2+ (aq) + CO 3 2- (aq) What is the “K-equation”? K = [Fe 2+ ][CO 3 2- ] The “K” is the PRODUCT of the SOLUBLE ions. Hence, this reaction is called a “solubility product”. K sp = [Fe 2+ ][CO 3 2- ] K sp (FeCO 3 ) = 3.07x TEXT AND YOUR QUESTION TO 37607

What is the solubility of FeCO 3 ? K sp (FeCO 3 ) = 3.07x FeCO 3 (s)  Fe 2+ (aq) + CO 3 2- (aq) IS00 C-x+x+x E xx K sp = 3.07x = [Fe 2+ ][CO 3 2- ] = x*x x = SQRT(3.07x ) = 5.54x10 -6 M TEXT AND YOUR QUESTION TO 37607

Clicker question What is the solubility of Ba 3 (PO 4 ) 2 at 298K? K sp (Ba 3 (PO 4 ) 2 ) = 6x A. 8x M B. 2x10 -8 M C. 3x10 -9 M D. 9x10 -9 M E. 3x M TEXT AND YOUR QUESTION TO 37607

Ba 3 (PO 4 ) 2 (s) 3 Ba 2+ (aq) + 2 PO 4 3- (aq) I S 00 C -x+3x+2x E -3x2x K sp = 6x = (3x) 3 (2x) 2 = 27x 3 *4x x = x 5 x = 8.89x10 -9 M = 9x10 -9 M TEXT AND YOUR QUESTION TO 37607

More common units for solubility… …are g/L. If you wanted g/L TEXT AND YOUR QUESTION TO 37607

Precipitation Reaction The reverse reaction: Solubility Product: Ba 3 (PO 4 ) 2 (s) 3 Ba 2+ (aq) + 2 PO 4 3- (aq) Precipitation: 3 Ba 2+ (aq) + 2 PO 4 3- (aq)  Ba 3 (PO 4 ) 2 (s) It’s just K “upside down” TEXT AND YOUR QUESTION TO 37607

How do you know if something precipitates? Ba 3 (PO 4 ) 2 (s) 3 Ba 2+ (aq) + 2 PO 4 3- (aq) K sp = 6x What is the K sp ? It’s the limit on the amount of ions in solution. K sp = [Ba 2+ ] 3 [PO 4 3- ] 2 Remember our old friend “Q”? TEXT AND YOUR QUESTION TO 37607

What’s Q? Q is just the concentrations of products and reactants when you are NOT at equilibrium. K sp = [Ba 2+ ] 3 [PO 4 3- ] 2 = 6x Q = [Ba 2+ ] 3 [PO 4 3- ] 2 = any other number TEXT AND YOUR QUESTION TO 37607

Q is less than K means… 1. You are NOT at equilibrium. 2. You could dissolve more solid: the products (dissolved ions) are too small. TEXT AND YOUR QUESTION TO 37607

Q is more than K means… 1. You are NOT at equilibrium. 2. You have TOO MANY products (dissolved ions). They can’t stay dissolved, they need to precipitate out! TEXT AND YOUR QUESTION TO 37607

A little precipitation question: 500 mL of M Fe(NO 3 ) 3 is mixed with 250 mL of M KOH. What, if anything, precipitates from the solution? What mass of precipitate is formed? TEXT AND YOUR QUESTION TO 37607

What COULD form…? KOH(s)  K + (aq) + OH - (aq) Fe(NO 3 ) 3 (s) Fe 3+ (aq) + 3 NO 3 - (aq) A beaker of KOH and Fe(NO 3 ) 3 has neither KOH nor Fe(NO 3 ) 3, it’s all ions! TEXT AND YOUR QUESTION TO 37607

The 1 st Rule of Chemistry… Opposites attract! Positive ions like negative ions. Negative ions like positive ions. Postive ions hate positive ions. Negative ions hate negative ions. Fe 3+ OH - NO 3 - K+K+ TEXT AND YOUR QUESTION TO 37607

Only possible products are… KOH or KNO 3 Fe(OH) 3 or Fe(NO 3 ) 3 We know that KOH and Fe(NO 3 ) 3 don’t form…that’s what we started with. What about KNO 3 and Fe(OH) 3 ? Fe 3+ OH - NO 3 - K+K+ TEXT AND YOUR QUESTION TO 37607

How many “Ks” are there in the beaker? A. None of the below B. 3 C. 2 D. 4 E. 5 TEXT AND YOUR QUESTION TO 37607

What about KNO 3 and Fe(OH) 3 ? They are both possible products of the reaction. Could they both form? Which one forms first? Do they form together? How would you know? K sp TEXT AND YOUR QUESTION TO 37607

When you have 2 possible reactions… BIGGEST K wins! Or, in this case, SMALLEST K sp K precipitation = 1/K sp Small K sp means big K precipitation. TEXT AND YOUR QUESTION TO 37607

Precipitation is just the reverse of dissolution. KNO 3 (s) K + (aq) + NO 3 - (aq) K sp (KNO 3 ) = HUGE (K + salts are very soluble and nitrates are very soluble) Fe(OH) 3 (s)  Fe 3+ (aq) + 3 OH - (aq) K sp (Fe(OH) 3 )=2.79x TEXT AND YOUR QUESTION TO 37607

So the only reaction to consider is… Fe(OH) 3 (s)  Fe 3+ (aq) + 3 OH - (aq) K sp (Fe(OH) 3 )=2.79x All equilibrium problems have 3 parts…yada yada yada… TEXT AND YOUR QUESTION TO 37607

K sp (Fe(OH) 3 )=2.79x Fe(OH) 3 (s)  Fe 3+ (aq) + 3 OH - (aq) I C E K sp = 2.79x = [Fe 3+ ][OH - ] 3 What do we know? TEXT AND YOUR QUESTION TO 37607

Don’t forget the dilution 500 mL of M Fe(NO 3 ) 3 is mixed with 250 mL of M KOH. So… Dilution is the solution! M x L = 0.05 mol/0.750 L = M M x L = mol/0.750 L = M TEXT AND YOUR QUESTION TO 37607

K sp (Fe(OH) 3 )=2.79x Fe(OH) 3 (s)  Fe 3+ (aq) + 3 OH - (aq) I C +x -x -3x E x x x K sp = 2.79x = [0.067-x][ x] 3 This is an algebraic mess BUT…K is really small. TEXT AND YOUR QUESTION TO 37607

K is really small… …which means… Fe(OH) 3 is not very soluble. So, x is going to be huge! We can use that to our advantage. We can mathematically precipitate out ALL of the Fe(OH) 3 and then redissolve it! TEXT AND YOUR QUESTION TO 37607

What is product limiting? Fe(OH) 3 (s)  Fe 3+ (aq) + 3 OH - (aq) I C - -x -3x E x x x = 0 X = – 3x = 0 X=0.011 The hydroxide runs out first! TEXT AND YOUR QUESTION TO 37607

What is product limiting? Fe(OH) 3 (s)  Fe 3+ (aq) + 3 OH - (aq) I C (0.011) E x = 0 X = – 3x = 0 X=0.011 The hydroxide runs out first! TEXT AND YOUR QUESTION TO 37607

Double your ICE, double your pleasure! Fe(OH) 3 (s)  Fe 3+ (aq) + 3 OH - (aq) I C *(0.011) I C -x +x +3x E x x 3x K sp = 2.79x = [0.056+x][3x] 3 Look how much simpler that is. Even better, let’s try and solve it the easy way! TEXT AND YOUR QUESTION TO 37607

Double your ICE, double your pleasure! K sp = 2.79x = [0.056+x][3x] 3 Assume x<<0.056! 2.79x = [0.056][3x] 3 = 0.056*27x x = x x = x! Pretty good assumption. TEXT AND YOUR QUESTION TO 37607

Double your ICE, double your pleasure! Fe(OH) 3 (s)  Fe 3+ (aq) + 3 OH - (aq) I C *(0.011) I C x x (1.23x ) E x M Fe(OH) 3 precipitate M Fe(OH) 3 *0.75 L * g/mol = 0.9 g Fe(OH) 3 TEXT AND YOUR QUESTION TO 37607

Neat trick, huh? Actually, that is another little trick in your ICE arsenal… We know what to do when x is small. Now, if we suspect x is large, we can try this little trick. In fact, you could always forcibly do a reaction to change the initial condition. After all, in the end the equilibrium will decide where it finishes. TEXT AND YOUR QUESTION TO 37607

Write a balanced equation for the dissolution of FeCO 3 (s) in water. A. FeCO 3 (s) ↔ Fe 3+ (aq) + CO 3 3- (aq) B. FeCO 3 (s) + H 2 O (l) ↔ Fe 2+ (aq) + CO 3 2- (aq) C. FeCO 3 (s) ↔ Fe 3+ (aq) + CO 3 2- (aq) D. FeCO 3 (s) ↔ Fe 2+ (aq) + CO 3 2- (aq) E. FeCO 3 (s) ↔ Fe (aq) + CO 3 (aq) TEXT AND YOUR QUESTION TO 37607

Write a balanced equation for the dissolution of MgCO 3 (s) in water. A. MgCO 3 (s) ↔ Mg 3+ (aq) + CO 3 3- (aq) B. MgCO 3 (s) + H 2 O (l) ↔ Mg 2+ (aq) + CO 3 2- (aq) C. MgCO 3 (s) ↔ Mg 2+ (aq) + CO 3 2- (aq) D. MgCO 3 (s) → Mg 2+ (aq) + CO 3 2- (aq) E. MgCO 3 (s) ↔ Mg (aq) + CO 3 (aq) TEXT AND YOUR QUESTION TO 37607

Write a balanced equation for the dissolution of Mg(NO 3 ) 2 (s) in water. A. Mg(NO 3 ) 2 (s) ↔ Mg 2+ (aq) + NO 3 2- (aq) B. Mg(NO 3 ) 2 (s) ↔ Mg + (aq) + NO 3 - (aq) C. Mg(NO 3 ) 2 (s) ↔ Mg 2+ (aq) + 2 NO 3 - (aq) D. Mg(NO 3 ) 2 (s) ↔ Mg + (aq) + 2 NO 3 - (aq) E. Mg(NO 3 ) 2 (s) ↔ Mg 2+ (aq) + 2 NO 3 2- (aq) TEXT AND YOUR QUESTION TO 37607

Write a balanced equation for the dissolution of Fe(NO 3 ) 2 (s) in water. A. Fe(NO 3 ) 2 (s) ↔ Fe 2+ (aq) + NO 3 2- (aq) B. Fe(NO 3 ) 2 (s) ↔ Fe + (aq) + NO 3 - (aq) C. Fe(NO 3 ) 2 (s) ↔ Fe 2+ (aq) + 2 NO 3 2- (aq) D. Fe(NO 3 ) 2 (s) ↔ Fe + (aq) + 2 NO 3 - (aq) E. Fe(NO 3 ) 2 (s) ↔ Fe 2+ (aq) + 2 NO 3 - (aq) TEXT AND YOUR QUESTION TO 37607

Write a balanced equation for the dissolution of K 2 CO 3 (s) in water. A. K 2 CO 3 (s) ↔ 2 K + (aq) + CO 3 2- (aq) B. K 2 CO 3 (s) ↔ K 2+ (aq) + CO 3 2- (aq) C. K 2 CO 3 (s) ↔ 2K + (aq) + CO 3 - (aq) D. K 2 CO 3 (s) ↔ K + (aq) + CO 3 2- (aq) E. K 2 CO 3 (s) ↔ K (aq) + CO 3 (aq) TEXT AND YOUR QUESTION TO 37607

If I dissolve Mg(NO 3 ) 2, Fe(NO 3 ) 2, and K 2 CO 3 in water, what species are present in my beaker? A. Mg, NO 3, Fe, NO 3, K, CO 3, H 2 O B. Mg 2+, NO 3 -, Fe 3+, NO 3 -, K 2+, CO 3 2-, H 2 O C. Mg 2+, Fe 3+, NO 3 -, K +, CO 3 2-, H 2 O D. Mg 2+, Fe 2+, NO 3 -, K +, CO 3 2-, H 2 O E. Mg 2+, Fe 3+, CO 3 2-, H 2 O TEXT AND YOUR QUESTION TO 37607

Given the following Ksp values, which salt is LEAST soluble in water? A. K sp [FeCO 3 ] = 3.07x B. K sp [MgCO 3 ] = 6.82x10 -6 C. K sp [Mg(NO 3 ) 2 ] = 3.14x10 4 D. K sp [Fe(NO 3 ) 2 ] = 6.28x10 3 E. K sp [K 2 CO 3 ] = 1.26x10 5 TEXT AND YOUR QUESTION TO 37607

Be a little careful about stoichiometry… Imagine the following chloride salts: K sp (XCl 2 ) = 12 K sp (YCl) = 9 Which salt is MORE SOLUBLE? A. XCl 2 B. YCl C. I need more information TEXT AND YOUR QUESTION TO 37607

A. I did Problem 8 B. I did not get to Problem 8 yet C. I don’t know how to do Problem 8 TEXT AND YOUR QUESTION TO 37607

What precipitates? A. K 2 CO 3 B. Mg(NO 3 ) 2 C. Fe(NO 3 ) 2 D. FeCO 3 E. MgCO 3 TEXT AND YOUR QUESTION TO 37607

Consider a solution that is 2.3×10 -2 M in Fe(NO 3 ) 2 and 1.5×10 -2 M in Mg(NO 3 ) 2. What must the concentration of potassium carbonate be to get the iron ions to start to precipitate? A. None – the iron ions won’t precipitate. B M C. 5.5x10 -6 M D. 1.3x10 -9 M E M TEXT AND YOUR QUESTION TO 37607

Consider a solution that is 2.3×10 -2 M in Fe(NO 3 ) 2 and 1.5×10 -2 M in Mg(NO 3 ) 2. What must the concentration of potassium carbonate be to get the magnesium ions to start to precipitate? A M B. 2.1x10 6 M C. 4.5x10 -4 M D M E. 723 M TEXT AND YOUR QUESTION TO 37607

What minimum concentration of K 2 CO 3 is required to cause the precipitation of the cation that precipitates first? A. 2.1x10 6 M B. 4.5x10 -4 M C. 5.5x10 -6 M D. 1.3x10 -9 M E M TEXT AND YOUR QUESTION TO 37607

Puzzle 1. How much of the first cation is still in solution when the second cation begins to precipitate? A. 4.55x10 -4 M B. 6.75x10 -8 M C. 1.5x10 -2 M D. 1.3x10 -2 M E. 2.3x10 -2 M TEXT AND YOUR QUESTION TO 37607

Puzzle 2. What is the total mass of K 2 CO 3 that must be added to get the second cation to begin to precipitate? A g B g C g D g E g TEXT AND YOUR QUESTION TO 37607

Clicker Questions What affect would adding acid have on the solubility of Ca(OH) 2 ? A. Increase the solubility B. Decrease the solubility C. Have no effect on the solubility. TEXT AND YOUR QUESTION TO 37607

The solubility of FeBr 2 in pure water is 1.2 g/100 mL at 298 K. Would you expect the solubility of FeBr 2 to be _______ in M NaBr at 298 K? A. higher B. lower C. the same TEXT AND YOUR QUESTION TO 37607

LeChatelier’s principle If you stress an equilibrium, the equilibrium shift to respond to the stress. Ca(OH) 2 (s) → Ca 2+ (aq) + 2 OH - (aq) K sp = [Ca 2+ ][OH - ] 2 “Lowering the pH” means increasing [H 3 O + ] which will neutralize some of the [OH-]. If you decrease the amount of hydroxide the reaction needs to make more to keep it at equilibrium. TEXT AND YOUR QUESTION TO 37607

REACTIONS ARE STUPID! They don’t know where the Ca 2+ or the OH - come from. TEXT AND YOUR QUESTION TO 37607

LeChatelier’s Principle Adding NaBr gives you a second source of Br-. So the reaction needs to make less to get to equilibrium. NaBr(s) → Na + (aq) + Br - (aq) K sp =[Na + ][Br - ] Still stupid! TEXT AND YOUR QUESTION TO 37607

Sample question If you have mL of a solution that is M in Fe 2+ and M in Mg 2+ and add mL of M K 2 CO 3. What is left in solution after the precipitation? K sp (FeCO 3 ) = 3.07x K sp (MgCO 3 ) = 6.82x10 -6 TEXT AND YOUR QUESTION TO 37607

K sp (FeCO 3 ) = 3.07x K sp (MgCO 3 ) = 6.82x10 -6 I expect the FeCO 3 to precipitate first TEXT AND YOUR QUESTION TO 37607

FeCO 3 (s) ↔ Fe 2+ (aq) + CO 3 2- (aq) 3.07x = ( x)( x) Assume x is LARGE! x x x TEXT AND YOUR QUESTION TO 37607

FeCO 3 (s) ↔ Fe 2+ (aq) + CO 3 2- (aq) 3.07x = ( x)(x) Assume x << x = x X=1.56x x+x x xx TEXT AND YOUR QUESTION TO 37607

FeCO 3 (s) ↔ Fe 2+ (aq) + CO 3 2- (aq) “M ” “M” x x “M” x10 -9 TEXT AND YOUR QUESTION TO 37607

Check MgCO 3 equilibrium After the FeCO 3 precipitates, the CO 3 2- concentration is only 1.56x10 -9 K sp (MgCO 3 ) = 6.82x10 -6 Q sp = (1.56x10 -9 )( M)=2.137x Q << K, so no MgCO 3 precipitates! TEXT AND YOUR QUESTION TO 37607

In terms of the solid… FeCO 3 (s) ↔ Fe 2+ (aq) + CO 3 2- (aq) Effectively “ M” FeCO 3 precipitated. This is not a real concentration – it’s not dissolved anymore! “M ” “M” x x “M” x10 -9 TEXT AND YOUR QUESTION TO 37607

In terms of the solid: TEXT AND YOUR QUESTION TO 37607

Let’s take it a little farther. Suppose we add another mL of M K 2 CO 3 ? Well, we do the same thing all over again but we are starting with less Fe 2+ and there’s some initial CO 3 2- still floating around. The reaction starts where the previous one left off! TEXT AND YOUR QUESTION TO 37607

Here’s where we left off… FeCO 3 (s) ↔ Fe 2+ (aq) + CO 3 2- (aq) So… And we still have g of FeCO 3 in the bottom of the beaker! “M ” “M” x x “M” x10 -9 TEXT AND YOUR QUESTION TO 37607

We need to do the DILUTION! TEXT AND YOUR QUESTION TO 37607

Here’s where we start… FeCO 3 (s) ↔ Fe 2+ (aq) + CO 3 2- (aq) Same problem as before – x is BIG! Same solution as before – precipitate it all and then let it redissolve! x x x TEXT AND YOUR QUESTION TO 37607

CO 3 2- is thelimiting reactant FeCO 3 (s) ↔ Fe 2+ (aq) + CO 3 2- (aq) 3.07x = ( x)(x) Assume x << x = x x=1.775x10 -9 M x+x x xX TEXT AND YOUR QUESTION TO 37607

So, we have made more FeCO 3 FeCO 3 (s) ↔ Fe 2+ (aq) + CO 3 2- (aq) Again, it’s sort of a fake Molarity for the solid x10 -9 M+1.775x10 -9 M “M” x10 -9 M TEXT AND YOUR QUESTION TO 37607

In terms of the solid: TEXT AND YOUR QUESTION TO 37607

Check MgCO 3 equilibrium After the FeCO 3 precipitates, the CO 3 2- concentration is only 1.56x10 -9 K sp (MgCO 3 ) = 6.82x10 -6 Q sp = (1.775x10 -9 )( M)=2.386x Q << K, so no MgCO 3 precipitates! TEXT AND YOUR QUESTION TO 37607

If I compare my two results After 1 st addition:  g FeCO 3  M Fe 2+ 1.56x10 -9 M CO 3 2-  M Mg 2+ After 2 nd addition:  g FeCO 3  M Fe 2+ 1.775x10 -9 M CO 3 2-  M Mg 2+ Almost all the CO 3 2- precipitates each time. I’ve doubled the amount of FeCO 3. The Fe 2+ is dropping and the CO 3 2- is gradually increasing. The Mg 2+ is being diluted but the actual amount dissolved isn’t changing You could keep doing this until virtually all the FeCO 3 is gone EXCEPT there is also Mg 2+ in solution. TEXT AND YOUR QUESTION TO 37607

What happens to the Mg 2+ ? The CO 3 2- is creeping upwards. Eventually, you’ll get to the point at which it precipitates! When…? Check K sp !!! (Note: There is a dilution issue, so we’ll use the undiluted values to get a ballpark figure.) K sp (MgCO 3 ) = 6.82x10 -6 =[Mg 2+ ][CO 3 2- ] TEXT AND YOUR QUESTION TO 37607

What happens to the Mg 2+ ? At all times, BOTH equilibria are in play IF the concentrations are high enough. At first only the FeCO 3 equilibrium is an issue because there’s not enough CO 3 2- to exceed the MgCO 3 K sp. Once it kicks in, however, BOTH equilibria must be satisfied all the time. So, when is this…? K sp (MgCO 3 ) = 6.82x10 -6 =[Mg 2+ ][CO 3 2- ] 6.82x10 -6 =[0.014 M][CO 3 2- ] [CO 3 2- ]=4.87x10 -4 M So, until the CO 3 2- concentration has increased to about 5x10 -4 M, the Mg 2+ won’t do anything! Eventually, it will get there… TEXT AND YOUR QUESTION TO 37607

If I compare my two results After 1 st addition:  g FeCO 3  M Fe 2+ 1.56x10 -9 M CO 3 2-  M Mg 2+ After 2 nd addition:  g FeCO 3  M Fe 2+ 1.775x10 -9 M CO 3 2-  M Mg 2+ Almost all the CO 3 2- precipitates each time. I’ve doubled the amount of FeCO 3. The Fe 2+ is dropping and the CO 3 2- is gradually increasing. The Mg 2+ is being diluted but the actual amount dissolved isn’t changing You could keep doing this until virtually all the FeCO 3 is gone EXCEPT there is also Mg 2+ in solution. TEXT AND YOUR QUESTION TO 37607

When…? Depends on the FeCO 3 equilibrium! When the Fe 2+ concentration has dropped enough to allow the CO 3 2- concentration to get high enough! K sp (FeCO 3 ) = 3.07x =[Fe 2+ ][4.87x10 -4 ] [Fe 2+ ]=6.304x10 -8 M So, we need to precipitate enough Fe 2+ to drop the initiall 0.22 M Fe 2+ down to 6.304x10 -8 M (ignoring dilution) TEXT AND YOUR QUESTION TO 37607

In terms of amounts: TEXT AND YOUR QUESTION TO 37607

Once the Mg 2+ starts precipitating… There is still a wee bit of Fe 2+ left that will co-precipitate with the Mg 2+. There ain’t a ton, so you are precipitating mostly pure Mg 2+ in this case. But that depends on how different the K sp is. If they were closer together, you might have more left. Only the first compound to precipitate will precipitate in pure form! TEXT AND YOUR QUESTION TO 37607