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## ICE

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**ICE**• The equilibrium constant can be calculated using only one equilibrium concentration , if the starting concentrations of the substances are known. This technique uses our knowledge of equilibrium formation to determine the concentrations of the reactant and products at equilibrium. The technique uses an ICE chart to solve the calculation. ICE stands for Initial, Change, Equilibrium.**Keq**• Steps to Solving Problems: • Write an equilibrium expression for the balanced reaction. • Write an ICE table. Fill in the given amounts. • Use stoichiometry (mole ratios) on the change in concentration line. • Determine the equilibrium concentrations of all species. • Usually, the initial concentration of products is zero. (This is not always the case.)**Given only one equilibrium concentration and starting**concentrations (ICE) 2 HI (g) H2 (g) + I2 (g) Keq = 1.26 *10-3 • The decomposition of HI was studied by injecting 2. 50 mole of HI into at 10.32 L vessel at 25oC. What is the [H2] at equilibrium?**ICE - Predicting final concentrations**• Calculate the initial and equilibrium concentration for chemical substance in question. • If initial concentration is not listed it can be assumed to be zero • Remember, the square brackets mean concentration, in molarity, if it is given in some other unit you must calculate the molarity.**Given only one equilibrium concentration and starting**concentrations (ICE) • Create an ICE chart to hold calculations**2 HI (g) H2 (g) + I2 (g) Keq = 1.26 *10-3**The decomposition of HI was studied by injecting 2.50 mole of HI into at 10.32 L vessel at 25oC. What is the [H2] at equilibrium? I2 HI H2**The decomposition of HI was studied by injecting 2.50 mole**of HI into at 10.32 L vessel at 25oC. What is the [H2] at equilibrium? Note the moles into a 10.32 L vessel stuff - calculate molarity. Starting concentration of HI: 2.50 mol/10.32 L = 0.242 M And we know the starting [ ] of H2 and I2 must be 0 0 0 0.242M**The decomposition of HI was studied by injecting 2.05 mole**of HI into at 10.32 L vessel at 25oC. What is the [H2] at equilibrium? Write the Kc equation [H2] [I2] =1.26 *10-3 Ksp= [HI]2**The decomposition of HI was studied by injecting 2.50 mole**of HI into at 10.32 L vessel at 25oC. What is the [H2] at equilibrium? 2HI (g) H2 (g) + I2 (g) Kc = 1.26 *10-3 We are looking for [H2] or x From the equation we know that the [H2] is equal to [I2] is equal to 2[HI] x x -2 x**The decomposition of HI was studied by injecting 2.50 mole**of HI into at 10. 32 L vessel at 25oC. What is the [H2] at equilibrium? • Finish the chart 0 + x 0.242-2x 0 + x**[H2] [I2]**= 1.26 *10-3 Kc = [HI]2 • The decomposition of HI was studied by injecting 2.50 mole of HI into at 10. 32 L vessel at 25oC. What is the [H2] at equilibrium? Now fill in the Kc equation with the information you have**[H2] [I2]**= 1.26 *10-3 Kc = [HI]2 Kc = [0+x] [0+x] = 1.26*10-3 [0.242-2x]2**Keq = [0+x] [0+x]**=1.26*10-3 [0.242-2x]2 Keq = x2 =1.26*10-3 [0.242-2x]2**And yes, it’s a quadratic equation. Do a bit of**rearranging x2 =1.26*10-3 Keq = [0.242-2x]2 x2 = 1.26*10-3 * [0.242-2x]2 x2 = 1.26*10-3 * [0.0586 - 0.968x + 4x2] x2 = 7.38*10-5 – 1.22 *10-3 x + 5.04 *10-3 x2 0.995x2 +1.22*10-3 x - 7.38*10-5 = 0**X = -b + b2 -4ac**2a 0.995x2 +1.22*10-3 x - 7.38*10-5 = 0 X= -1.22*10-3 + (1.22*10-3)2 -4(0.995* 7.38*10-5) 2 (0.995) x = 0.00804 or –0.00925 Since we can’t have – concentrations we know the [H2 ] is 0.00804 M.**Solubility Product Constant**The solubility of a solute is the maximum quantity of solute that can dissolve in a certain quantity of solvent or quantity of solution at a specified temperature. Generally expressed in two ways: grams of solute per 100 g of water moles of solute per Liter of solution**Solubility Product Constant**• A saturated solution is in a state of dynamic equilibrium between the dissolved, dissociated, ionic compound and the undissolved solid. The equation is written: MxAy(s) x M y+ (aq) + y Ax-(aq)**Solubility Product Constant**MxAy(s) x M +y (aq) + y A-x (aq) Now write the Keq for this equations: [M+y] x y [A-x] = [M+y]x [A-x]y Ksp= [MxAy(s)]**Solubility Product Constant**[My+]x [Ax-]y Ksp= Since the equilibrium constant refers to the product of the concentration of the ions that are present in a saturated solution of an ionic compound, it is given the name solubility product constant, and given the symbol Ksp.**Calculate the concentration of Ag+ andBr- ion in a saturated**solution. The Ksp of AgBr is 5.0*10-13 AgBr(s) Ag+(aq)+Br-(aq) Ksp = [Ag+][Br-] = 5.0 * 10-13 Ksp = x*x =5.0 * 10-13 or X2= 5.0 * 10-13 Or X= 7.1 * 10-7M H2O**X= 7.1 * 10-7M**Once we know how many moles of AgBr dissolve in a liter of water, we can calculate the solubility in grams per liter. 7.1 x 10-7mole AgBr | 187.8g AgBr = 1L | 1 mole AgBr 1.3*10-4 g/1L**When an excess of a slightly soluble ionic compound is mixed**with water, an equilibrium is established between the solid and the ions in the saturated solution. • For the salt calcium oxalate, CaC2O4, you have the following equilibrium. • H2O • CaC2O4 (s) Ca 2+ (aq) + C2O4 2- (aq)**Calculating Ksp from the Solubility**• A 1.0-L sample of a saturated calcium oxalate solution, CaC2O4, contains 0.0061g of the salt at 25 oC. Calculate the Ksp for this salt at 25 oC. • First convert the solubility of calcium oxalate from 0.0061 g/liter to moles per liter. M CaC2O4= 0.0061g CaC2O4 |1 mol CaC2O4 1L| 128 gCaC2O4 M =4.8*10-5 mol CaC2O4/1L**Calculating Ksp from the Solubility**• A 1.0-L sample of a saturated calcium oxalate solution, CaC2O4, contains 0.0061 g of the salt at 25 oC. Calculate the Ksp for this salt at 25 oC. • CaC2O4 (s) Ca 2+ (aq) + C2O4 2- (aq) • H2O • When 4.8 x 10-5 mol of solid dissolve it forms 4.8 x 10-5 mol of each ion. 0 0 +4.8*10 -5 + 4.8*10 -5 4.8*10 -5 4.8*10 -5**Calculating Ksp from the Solubility**• You can now substitute into the equilibrium-constant expression. Ksp =[Ca 2+ ] [C2O42-] Ksp =[4.8*10 -5 ] [4.8*10 -5 ] Ksp = 2.3*10-9**Calculating Ksp from the Solubility**• By experiment, you found that 1.2 x 10-3 mol of lead(II) iodide, PbI2, dissolves in 1.0 L of water at 25 oC. What is the Ksp at this temperature? • Note - in this example, you find that 1.2 x 10-3 mol of the solid dissolves to give 1.2 x 10-3 mol Pb2+ and (2 x (1.2 x 10-3)) mol of I-. • PbI2 Pb 2+ (aq) + 2I- (aq)**H2O**• PbI2(s) Pb 2+ (aq) + 2I- (aq) Calculating Ksp from the Solubility • By experiment, you found that 1.2 x 10-3 mol of lead(II) iodide, PbI2, dissolves in 1.0 L of water at 25 oC. What is the Ksp at this temperature? 0 0 + 1.2 x 10-3 +2 *(1.2 x 10-3) + 1.2 x 10-3 + 2.4 x 10-3**Calculating Ksp from the Solubility**• By experiment, it is found that 1.2 x 10-3 mol of lead(II) iodide, PbI2, dissolves in 1.0 L of water at 25 oC. What is the Ksp at this temperature? • Substituting into the equilibrium-constant expression: Ksp =[Pb 2+ ] [I -]2 Ksp =[1.2 x 10-3] [2.4 x 10-3]2 Ksp = 6.9*10-9**Calculating the Solubility from Ksp**• Your turn • The mineral fluorite is calcium fluoride, CaF2. Calculate the solubility (in grams per liter) of calcium fluoride in water from the Ksp (3.4 x 10-11)**Calculating the Solubility from Ksp**• The mineral fluorite is calcium fluoride, CaF2. Calculate the solubility of calcium fluoride (in grams per liter) in water from the Ksp (3.4 x 10-11) • Write the Ksp equation • Set up the table • Solve for x**Calculating the Solubility from Ksp**• The mineral fluorite is calcium fluoride, CaF2. Calculate the solubility (in grams per liter) of calcium fluoride in water from the Ksp (3.4 x 10-11) • Convert to g/L • (CaF2 78.1 g/mol). 1.6 *10-2 gCaF2 /L 2.0*10-4 mol CaF2 | 78.1 g CaF2 = 1L | 1 mole CaF2**Solubility and the Common-Ion Effect**• In this section we will look at calculating solubilities in the presence of other ions. • The importance of the Ksp becomes apparent when you consider the solubility of one salt in the solution of another having the same cation.**Solubility and the Common-Ion Effect**• In this section we will look at calculating solubilities in the presence of other ions. • For example, suppose you wish to know the solubility of calcium oxalate in a solution of calcium chloride. • Each salt contributes the same cation (Ca2+) • The effect is to make calcium oxalate less soluble than it would be in pure water.**What is the molar solubility of calcium oxalate in 0.15 M**calcium chloride? The Ksp for calcium oxalate is 2.3*10-9. • CaC2O4 (s) Ca 2+ (aq) + C2O4 2- (aq) • H2O • Note that before the calcium oxalate dissolves, there is already 0.15 M Ca2+ in the solution.. 0 0.15 x x x 0.15 +x**You substitute into the equilibrium-constant equation**• What is the molar solubility of calcium oxalate in 0.15 M calcium chloride? The Ksp for calcium oxalate is 2.3 * 10-9. Ksp =[Ca 2+ ] [C2O42-] Ksp = [0.15 +x] [x] = 2.3*10-9 • Now rearrange this equation to give = [x] = 2.3*10-9 [0.15 +x] 2.3*10-9 [0.15] 1.5 *10-8 But since we expect x to be very very small compared to 0.15, we do this**What is the molar solubility of calcium oxalate in 0.15 M**calcium chloride? The Ksp for calcium oxalate is 2.3 x 10-9. • Therefore, the molar solubility of calcium oxalate in 0.15 M CaCl2 is 1.5 x 10-8 M. • In pure water, the molarity was 4.8 x 10-5 M, which is over 3000 times greater.**Predicting Precipitation**• Predicting whether precipitation occurs boils down to one question. Is my ION PRODUCT (Kip) greater than the KNOWN SOLUBILITY PRODUCT (KSP)? Kip > Ksp then precipitation. Ksp > Kip no precipitation.**Determining whether a precipitate will, or will not form**when two solutions are combined • When two electrolytic solutions are combined, a precipitate may, or may not form. In order to determine whether or not a precipitate will form or not, one look at two factors. • First, determine the possible combinations of ions that could result when the two solutions are combined to see if any of them are deemed "insoluble" base on solubility tables (Ksp tables will also do). • Second, determine if the concentrations of the ions are great enough so that the reaction quotient Q exceeds the Ksp value. One important factor to remember is there is a dilution of all species present and must be taken into account.**Problem**• The final concentration of CaF2 in a solution is 1.5 x10-5 M and the final concentration of NaF in a solution AFTER mixing is 2.2 x10-2 M Will precipitation occur? Ksp of CaF2 is 4.0 x10-11**Write a balanced dissolution reaction.**Hint: if you're not sure what will form, read the question carefully, the known Ksp of the compound must be given. That is what might or might not form depending on the ION PRODUCT). Ex. CaF2 Ca 2+ + 2F-**Write a Ksp expression but call it Kip (it will have our**data in it, from the concentrations that we put together in a test tube.) Kip = [Ca2+] [F-]2**Substitute the ACTUAL VALUES of the concentrations of the**ions that we actually have. No X's because we have no unknowns. We just want to see if our ion product is greater than the known Ksp. If so, the solution can't hold that many ions and precipitation occurs. Ca 2+ = 1.5 x10-5 M F- = 2.2 x10-2 M Kip= [Ca 2+] [F-]2**Substitute**• Kip= [Ca 2+] [F-]2Kip = [1.5 *10-5 M] [2.2 *10-2 M]2 Kip= 7.26 *10-9 Ksp - from problem is 4.0*10-11 Kip is 180 x more than Ksp so we get a precipitate.**Lead chloride, PbCl2, has a Ksp value of**1.7 x 10-5. Will a precipitate form when 140.0 ml of 0.010 M Pb3(PO4)2 is mixed with 550.0 ml of 0.055 M NaCl? Justify your answer with calculations.**What are we looking for?**Will a precipitate of PbCl2 form? 1) Write the net ionic equation for PbCl2 1PbCl2 Pb+2 + 2Cl- 1) Write the Ksp equation for PbCl2 the Ksp = [ Pb+2] + [Cl-]2 = 1.7 * 10-5**Lead chloride, PbCl2, has a Ksp of 1.7 x 10-5.**Will a precipitate form when 140.0 ml of 0.010 M Pb3(PO4)2 is mixed with 550.0 ml of 0.055 M NaCl? Justify your answer with calculations. We know Kip > Ksp then precipitation. Ksp > Kip no precipitation. We also know Kip = [Pb+2] [Cl-]2**the Kip = [ Pb+2] + [Cl-]2 = 1.7 * 10-5**What information do we need to solve for Kip We need the [ Pb+2] + [Cl-] Go back to the problem and find the information**Will a precipitate form when 140.0 ml of 0.010 M Pb3(PO4)2**is mixed with 550.0 ml of 0.055 M NaCl? 0.010 M Pb3(PO4)2 Pb3(PO4)2 3Pb+2 + 2PO4-3 • 0.010M in .140L = 0.0014 mole of Pb3(PO4)2 has been added to the tt. • [Pb+2 ] = 0.0014mole /.690L (the total volume of the test tube (140 mL + 550 mL) = 0.02M * 3 (3 Pb+2 for every 1 Pb3(PO4)2) = • [Pb+2 ] = .006086M**Will a precipitate form when 140.0 ml of 0.010 M Pb3(PO4)2**is mixed with 550.0 ml of 0.055 M NaCl? 0.055 M NaCl NaCl Na+ + Cl- • 0.055M in .550L = .03025 mole of NaClhas been added to the tt. • [Na+ ] = 03025 mole /.690L (the total volume of the test tube (140 mL + 550 mL) = 0.0435M = • [Na+ ] = .0435 M Na+**the Kip = [ Pb+2] + [Cl-]2 = 1.7 * 10-5**We know the [ Pb+2] + [Cl-] Go solve the problem [ .006086] [0.0435] 2 = 1.15*10-5 1.7 * 10-5 <1.15*10-5 no ppt