Solubility Product. We can compare this equilibrium to the water ionization equilibrium.

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1 Solubility Product The concentrations of ions in saturated solutions have a relationship to one another somewhat like the relationship between the concentration of H 3 O + and OH - in water. Sodium Chloride Consider saturated sodium chloride solution. Quite a bit of sodium chloride can be dissolved in water, about 6 moles in one liter. That makes the concentration of both the sodium ion and the chloride ion about 6 M. What happens if we increase the concentration of Cl - by adding some concentrated hydrochloric acid (12M HCl)?. The saturated NaCl is in the test tube and the concentrated HCl is in the dropper. If nothing happened, we would still have 6M Na + and a higher concentration, perhaps, 8M, of Cl -. But something does happen. Crystals of NaCl form from the reaction of some of the extra Cl - with some of the Na + that was in the solution. The concentration of Na + goes down to around 5 M as the conc of Cl - increases to somewhere around 7M. Approximate Concentrations [Na + ] [Cl - ] Start 6M 6M Potential 6M 8M Actual 5M 7M As the concentration of one ion increases, the concentration of the other ion decreases. Just as there was an equation that related the concentrations of the dissociated ions of water, there is an equation that relates the concentrations of the dissociated ions of sodium chloride. We can compare this equilibrium to the water ionization equilibrium. Water ionizes to form H 3 O + and OH -. 2 H 2 O H 3 O + + OH - The reaction is reversible. 2 H 2 O H 3 O + + OH - The concentrations of H 3 O + and OH - are K related by this equation: w = [H 3 O + ] [OH - ] When multiplied together, the concentrations of H 3 O + and OH - give a fairly constant value called the ionization constant of water, or K w. Returning to sodium chloride we can see a similar relationship between the ions and solid (excess) as we have with the H 3 O + and OH - and the (excess) liquid water. Sodium chloride dissolves and dissociates in water to Na + and Cl -. NaCl(s) Na + + Cl - The reaction is reversible: NaCl(s) Na + + Cl -

2 The concentrations of Na + and Cl - are K realated by this equation: sp = [Na + ][Cl - ] When multiplied together, the concentrations of Na + and Cl - give a fairly constant value called the solubility product constant, or K sp. For sodium chloride, the value of K sp is about 36. Practice See if you can figure out what the concentration of Na + would be if we were able to increase the concentration of Cl - up to 10 M. Take a moment to figure that out. Answer You should have calculated about 3.6 M for the sodium ion concentration. I got that value by saying that the concentration of Na + times the concentration of Cl - is equal to 36 (the K sp value for sodium chloride). If the concentration of Cl - is going to be 10 M, then the concentration of of Na + has to be 36 divided by 10. That comes at to 3.6 M. K sp = [Na + ]x[cl - ] [Na + ] = K sp [Cl - ] [Na + ] = [Na + ] = 3.6M Silver Chloride The same line of reasoning can be used with any salt that dissolves in water, even if it dissolves only a very small amount. Silver chloride will dissolve somewhat in water. However, it reaches saturation very quickly--that is, when the concentrations of silver and chloride ions are about 1.3 x 10-5 M. Still we can write a solubility product equation for it. The value for the K sp of silver chloride, however, is about 1.8 x 10-10, a very small number. AgCl(s) Ag + + Cl - K sp = [Ag + ][Cl - ] K sp = 1.8 x 10-10

3 Practice Try using that information to calculate the Ag + concentration if the chloride ion concentration were 3.0 M. Answer In this case the answer turns out to be a very small number, which can be calculated using the process shown here. K sp = [Ag + ]x[cl - ] [Ag + ] = K sp [Cl - ] [Ag + ] = (1.8 x ) 3.0 [Ag + ] = 6.0 x M Lead(II) Chloride When the formula for a salt contains more than just one of each ion, the solubility product equation gets a little more complicated. Let's use PbCl 2 as an example. When PbCl 2 cissolved in water, two ions of Cl - are released for every one ion of Pb 2+. When saturation is reached we show that the reaction is reversible. PbCl 2 (s) Pb 2+ + Cl - PbCl 2 (s) Pb Cl - The equation for the solubility product is: K sp = [Pb 2+ ][Cl - ] 2 The concentration of Cl - is squared because the balanced equation for the reaction shows a 2 as the coefficient in front of Cl -. Looking at the reaction in this way might help you to remember that: PbCl 2 (s) Pb 2+ + Cl - + Cl - then: K sp = [Pb 2+ ][Cl - ][Cl - ] thus: K sp = [Pb 2+ ] [Cl - ] 2 Example - Determining concentrations at equilibrium Here is an example of how to calculate the equilibrium concentration of one substance given the K sp and the equilibrium concentrations of the other substance. For the reaction PbCl 2 (s) Pb +2 (aq) + 2 Cl - (aq), what is [Pb +2 ] at equilibrium K sp = [Pb +2 ] x [Cl] x 10-5 = [Pb +2 ] x (2.0 x 10-3 ) x 10-5 = [Pb +2 ] x 4.0 x 10-6

4 if [Cl - ] = 2.0 x 10-3 M given that the K sp = 2.0 x 10-5? 2.0 x x 10-6 = [Pb +2 ] 5.0 M = [Pb +2 ] Practice Problems: Equilibrium Concentrations The K sp for AgCl is 1.8 x If Ag + and Cl - are both in solution and in equilibrium with AgCl. What is [Ag + ] if [Cl - ] =.020 M? If Ag + and Cl - were both present at M, would a precipitate occur? What concentration of Ag + would be necessary to bring the concentration of Cl - to 1.0 x 10-6 M or lower? Answers: Equilibrium Concentrations Here are the answers to the questions above. The K sp for AgCl is 1.8 x If Ag + and Cl - are both in solution and in equilibrium with AgCl. What is [Ag + ] if [Cl - ] =.020 M? [Ag + ] = 9.0 x 10-9 M If Ag + and Cl - were both present at M, would a precipitate occur? Yes, a precipitate would occur because these concentrations together are higher than what the K sp allows. What concentration of Ag + would be necessary to bring the concentration of Cl - to 1.0 x 10-6 M or lower? [Ag + ] = 9.0 x 10-9 M or higher

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