Common Ion, Buffers, Titrations, Ksp (Ch. 15)

CHM 152 – Common Ion, Buffers, Titrations, Ksp (Ch. 15) Common Ion Effect 1. Calculate the pH of a 2.00 L solution containing 0.885 moles of hypochlorous acid (HClO) and 0.905 moles of NaClO. Given Ka for HClO is 3.0 x 10-8. What is in the beaker? A weak acid HClO, and its conjugate base, ClO- ions from NaClO. (Na+ ions are spectators) So we have a buffer and can use the buffer equation. So we need the concentrations of these in the beaker. HClO .885 moles / 2.00L = 0.4425M HClO (acid) NaClO .905 moles / 2.00L = 0.4525M NaClO that dissociates 100% so = 0.4525M ClO- (c. base) pH = 7.5229 + log( 0.4525 / 0.4425) = 7.53 (need 2 decimal places since Ka had two sig dig) 2. What is the pH of a solution containing 0.30 M NH3 and 0.15 M NH4NO3? Kb for NH3 = 1.8x10-5 NH3 is a weak base: NH3 + H2O NH4+ + OHNH4NO3 is a salt: NH4NO3  NH 4 + NO 3 ; thus NH 4 is a “common ion” NH4+ + OH- NH3 + H2O I C E [NH3] M 0.30 -x 0.30 - x [NH 4 ] M 0.15 +x 0.15 + x [H2O] [OH-] M 0 +x x  Kb = [ NH 4 ][OH  ] [ NH 3 ] Approximation: ignore –x, +x terms: 1.8x10-5 = 0.15x 0.30 x = [OH-] = 3.6x10-5 M pOH = -log 3.6x10-5 = 4.44 pH = 14 – 4.44 = 9.56 pH = 9.56 (This problem can also be solved using the Ka rxn: NH 4 NH3 + H+ ; if you use this reaction, you must convert Kb to its corresponding Ka value.) Buffer Solutions Give the formulas for two chemicals that would make a buffer solution in water. HF and KF 3. a) Calculate the pH if 5.50 grams nitric acid is added to a buffer system composed of 35.5 grams acetic acid and 32.4 grams lithium acetate in 2.00 liters of water. Note the small amount of nitric acid will not affect Page 1 of 15 the volume of 2.00 liters. b) What was the pH of the buffer system before the nitric acid addition? c) Explain the change, or lack of change, in pH after the addition of the nitric acid to the buffer system. b. I’m doing part b) first. Exactly the same as number 1 - we have a buffer. 35.5 g acetic acid ( mol / 60.0 g) = 0.5917 moles then / 2.00L = 0.2958M acid 32.4 g LiCH3COO ( mol / 65.9 g) = 0.4917 moles then/ 2.00L = 0.2458M which will dissociate and give 0.2458M acetate ion (c. base) pKa = 4.744 pH = 4.744 + log (0.2458 / 0.2958) = 4.66 a. Now add nitric acid to the buffer. The acetate ion will neutralize the nitric acid according to HNO3 + CH3COO-  NO3- + CH3COOH First figure out the nitric acid moles: 5.50 g ( mol / 63.0g) = 0.0873 moles moles acid: 35.5 g acetic acid ( mol / 60.0 g) = 0.5917 moles CH3COOH moles c. base: 32.4 g LiCH3COO ( mol / 65.9 g) = 0.4917 moles LiCH3COO which will equal moles CH3COO- since the salt is completely soluble Set up an Initial Final table. There is no equilibrium because nitric is strong. Reacts one way. HNO3 + CH3COO NO3+ CH3COOH 0.0873 moles 0.4917 moles 0 0.5917 moles all reacts, limiting -0.0873 +0.0873 +0.0873 0 0.4044 moles 0.0873 moles 0.6790 moles What is left is buffer solution, nitrate ion is a spectator. pH = 4.744 + log (0.4044 / 0.6790) = 4.52 (Note you could have divided the moles by 2.00 L but the RATIO is the same. c. The pH decrease only a small amount because the conjugate base (acetate ion CH3COO- ) of the buffer neutralized all the strong nitric acid. So after the reaction there was only weak acid and conjugate base left - a buffer - so the pH remained fairly constant. 4. Calculate the pH of a buffer solution containing 0.20 M HCHO2 and 0.30 M NaCHO2. The volume of the solution is 125 mL. Ka for HCHO2 =1.8x10-4 a) What is the pH of this buffer solution? Salt: NaCHO2  Na+ + CHO2pH = pKa + log (base / acid) = -log (Ka) + log (0.30 / 0.20) pH = 3.7447 + 0.17609 = 3.92 b) If 50.0 mL of 0.10 M NaOH is added to the buffer solution, what is the pH? (Notice that the volume of added base is significant in this problem. This requires diluted concentrations to be calculated.) Strong base: NaOH  Na+ + OHPage 2 of 15 diluted so recalculate M: M HCHO2 = M CHO2- = 0.30 M 125 ml  (175 ml ) = 0.21 M; 0.20 M 125 ml  (175 ml ) M OH- = = 0.14 M 0.10 M 50.0 ml  (175 ml ) = 0.029 M neutralization reaction: OH- + HCHO2  CHO2- + H2O Initial Change Final 0.029 -0.029 0 0.14 -0.029 0.11 0.21 +0.029 0.24 pH = pKa + log (base / acid) = 3.7447 + log (0.24 / 0.11) = 4.08 *For a buffer solution, pH only rises a little if a small amount of strong base is added. c) If 50.0 mL of 0.10 M HCl is added to the buffer solution, what is the pH? Strong acid: HCl  H+ + Cldiluted so recalculate M: M HCHO2 = M CHO2- = 0.30 M 125 ml  neutralization reaction: Initial Change Final (175 ml ) = 0.21 M; 0.20 M 125 ml  (175 ml ) M H+ = = 0.14 M 0.10 M 50.0 ml  = 0.029 M (175 ml ) H+ + CHO2-  HCHO2 0.029 -0.029 0 0.21 -0.029 0.18 0.14 +0.029 0.17 pH = pKa + log (base / acid) = 3.7447 + log (0.18 / 0.17) = 3.77 * For a buffer, pH only drops a little when a small amount of strong acid is added. Strong Acid-Strong Base Titrations 5. If it takes 54 mL of 0.10 M NaOH to neutralize 125 mL of an HCl solution, what is the concentration of the HCl? (Use solution stoichiometry – see chapter 3 in your textbook) 0.043 M HCl 6. If it takes 25 mL of 0.050 M HCl to neutralize 345 mL of NaOH solution, what is the concentration of the NaOH solution? Page 3 of 15 (Use solution stoichiometry – see chapter 3 in your textbook) 0.0036 M NaOH 7. If it takes 50 mL of 0.50 M KOH solution to completely neutralize 125 mL of sulfuric acid solution (H2SO4), what is the concentration of the H2SO4 solution? For problem 3, you need to divide your final answer by two, because H2SO4 is a diprotic acid, meaning that there are two acidic hydrogens that need to be neutralized during the titration. As a result, it takes twice as much base to neutralize it, making the concentration of the acid appear twice as large as it really is. 0.10 M H2SO4 8. Calculate the mass of NH3 needed to neutralize 30.00 mL of a 2.5 M solution of HNO3. First, determine the #moles of H+ from the 30 mL of HNO3 0.03 L x 2.5 mol/L HNO3 = 0.075 moles of H+ At the neutralization point (equivalence point): moles of H+ = moles of OHTherefore, 0.075 moles of NH3 are needed for complete neutralization of the H+ from the 30 mL of HNO3 NH3 molar mass = 17.03 g/mol 0.075 moles of NH3 X 17.03 g/mol = 1.3 grams of NH3 (2 sig figs) 9. If 1.25 grams of pure CaCO3 required 25.50 mL of a HCl solution for complete reaction, calculate the molarity of the HCl solution. Reaction: 2 HCl + CaCO3 → H2CO3 + CaCl2 CaCO3 molar mass = 100.09 g/mol 1.25 g CaCO3 / 100.09 g/mol = 0.01249 mol CaCO3 It takes 2 moles of HCl to react with 1 mole of CaCO3 and therefore 0.02498 mol of HCl must be in the 25.50 mL . Calculate molarity: 0.02498 mol HCl / 0.0255 L = 0.0980 (3 sig. figs) 10. How many mL of 0.500 M HCl are required to neutralize 35.4 mL of a 0.150 M NaOH solution? Moles of OH- contained in the 35.4 mL: 0.0354 L X 0.150 mol OH- /L = 0.00531 mol OH- At the neutralization point (equivalence point): moles of H+ = moles of OHSo we must calculate the volume of the 0.500 M HCl solution that contains 0.00531 moles of H + . Page 4 of 15 To calculate this: 0.00531 moles of H+ / (0.500 mol HCl / L) = 0.0106 L = 10.6 mL HCl (3 sig figs) 11. What volume of 0.49M KOH solution is needed to neutralize 840 mL of a 0.01M HNO 3 solution? Find the number of moles of H+ contained in 840 mL of the 0.01 M HNO3 solution. 0.840 L X (0.010 mol H+ / L) = 0.0084 mol H+ At the neutralization point (equivalence point): moles of H+ = moles of OHNow find the volume of the 0.49 M KOH solution that contains 0.0084 mol OH0.00840 mol OH- / (0.49 mol OH- / L) = 0.0171 L = 17 mL (2 sig figs) 12. Can I titrate a solution of unknown concentration with another solution of unknown concentration and still get a meaningful answer? Explain your answer in a few sentences. You cannot do a titration without knowing the molarity of at least one of the substances, because you’d then be solving one equation with two unknowns (the unknowns being M 1 and M2). 13. Explain the difference between an endpoint and equivalence point in a titration. Endpoint: When you actually stop doing the titration (usually, this is determined by a color change in an indicator or an indication of pH=7.0 on an electronic pH probe) Equivalence point: When the solution is exactly neutralized. It’s important to keep in mind that the equivalence point and the endpoint are not exactly the same because indicators don’t change color at exactly 7.0000 pH and pH probes aren’t infinitely accurate. Generally, you can measure the effectiveness of a titration by the closeness of the endpoint to the equivalence point. 14. Calculate the pH when 15.0 mL of 0.150M perchloric acid is added to 12.0 mL of 0.125M potassium hydroxide. Strong acid and strong base. Reacts one way. HClO4 (aq) + KOH (aq)  H2O (l) + KClO4 (aq) Need moles of each. acid: 0.0150L ( 0.150 mol / L ) = 0.00225 moles acid base: 0.0120L (0.125mol / L) = 0.00150 moles base Set up initial final table HClO4 + KOH  H2O (l) + KClO4 0.00225 moles 0.00150 moles --- 0 - 0.0015 all reacts, limiting --- + 0.0015 0.00075 moles 0 --- 0.00150 moles NOT a buffer by the way!!! KClO4 is a neutral salt, not a conjugate base. Note the new volume is 27.0 mL. pH will depend on the strong acid left over not the neutral salt. [H+] = 0.00075 moles / 0.0270L = 0.0278M pH = 1.56 (final answer needs 2 decimal places since 0.00075 moles had two sig dig) Page 5 of 15 15. Calculate the pH when 25.0 mL of 0.100M HBr is added to 15.0 mL of 0.100M LiOH. Strong acid and strong base. Reacts one way. HBr (aq) + LiOH (aq)  H2O (l) + LiBr (aq) Need mmoles of each. acid: 25.0 mL ( 0.100 mol / L ) = 2.50 mmoles acid base: 15.0 mL (0.100mol / L) = 1.50 mmoles base Determine how much acid is in excess: 2.50 mmol – 1.50 mmol = 1.00 mmol excess NOT a buffer by the way!!! LiBr is a neutral salt, not a conjugate base. Note the new volume is 40.0 mL. pH will depend on the strong acid left over not the neutral salt. So HBr dissociates 100%. Thus [H+] = 1.00 moles / 40.0 mL = 0.0250M pH = 1.602 (final answer needs 3 decimal places since everything had three sig figs) 16. How many mL of 0.225M barium hydroxide are needed to neutralize 20.0mL of 0.424M hydrobromic acid? Write the reaction and show each step in your stoichiometric calculation. Strong acid and strong base react completely. 2 HBr(aq) + Ba(OH)2(aq)  BaBr2(aq) + 2 H2O (l) (0.0200 L HBr)(0.424 mol / L)( 1 Ba(OH)2 / 2 HBr) ( L / 0.225 mol) (1000mL / L) = 18.8 mL Ba(OH)2(aq) 17. A 20.00 ml sample of 0.150 M HCl is titrated with 0.200 M NaOH. Calculate the pH of the solution after the following volumes of NaOH have been added: a) 0 mL; b) 10.00 mL; c) 15.0 mL; d) 20.00 mL. a) 0 ml of NaOH added – only SA is present initially: For strong acid: [H+] = [HCl] = 0.150 M HCl pH = -log[H+] = -log(0.150) = 0.824 b) 10.00 ml of NaOH neutralization reaction: HCl + NaOH  NaCl + H2O SA SB  1 L  0.150 moles HCl   moles HCl = 20.00 ml    3.00x10-3 moles HCl 1000 mL L      1 L  0.200 moles NaOH   moles NaOH = 10.00 ml    2.00x10-3 moles NaOH 1000 mL L    After neutralization: moles excess acid = 3.00x10-3 moles - 2.00x10-3 moles = 1.00x10-3 moles HCl M H+ = M HCl = 1.00 x10 3 moles  0.0333 M 0.03000 L Page 6 of 15 pH = - log [H+] = - log 0.0333 = 1.478 c) 15.0 mL of NaOH From part b, moles HCl = 3.00x10-3 moles HCl  1 L  0.200 moles NaOH   moles NaOH = 15.00 ml    3.00x10-3 moles NaOH L   1000 mL  moles HCl = moles NaOH at equivalence pt: pH = 7.000 (for SA/SB titration) d) 20.00 mL from part b, moles HCl = 3.00x10-3 moles HCl  1 L  0.200 moles NaOH   moles NaOH = 20.00 ml    4.00x10-3 moles NaOH 1000 mL L    After neutralization: moles excess base = 4.00x10-3 moles – 3.00x10-3moles = 1.00x10-3 moles NaOH 1.00 x10 3 moles  0.0250 M OH0.040 L pOH = -log 0.0250 = 1.602 pH = 14 – 1.602 = 12.398 M OH- = M NaOH = Weak Acid-Strong Base Titrations 18. A 50.0 mL sample of 0.500 M HC2H3O2 acid is titrated with 0.150 M NaOH. Ka = 1.8x10-5 for HC2H3O2. Calculate the pH of the solution after the following volumes of NaOH have been added: a) 0 mL; b) 166.7 mL; c) 180.0 mL. a) 0 ml of base; only a weak acid is initially present so [H+] ≠ [HA] HC2H3O2 I C E 0.500 -x 0.50-x H+ + C2H3O20 x x 0 x x  [ H  ][C 2 H 3 O2 ] Ka = [ HC2 H 3 O2 ] 1.8x10-5 = x2 0.500 [H +] = x = 0.500( 1.8 x10 5 ) = 3.0x10-3 pH = -log 3.0x10-3 = 2.52 b) 166.7 ml of NaOH are added  1 L  0.500 moles HC 2 H 3 O2  L  1000 mL  moles HC2H3O2 = 50.0 ml      2.50x10-2 moles HC2H3O2 Page 7 of 15  1 L  0.150 moles NaOH    L   1000 mL  moles NaOH = 166.7 ml  2.50x10-2 moles NaOH neutralization: HC2H3O2 + OH-  C2H3O2- + H2O I C Final 0.0250 -0.0250 0 0.0250 -0.0250 0 0 +0.0250 0.0250 only acetate remains – a weak base: [C2H3O2-] = 2.50 10 2 moles  0.2167 L 0.115 M base hydrolysis: C2H3O2- + H2O I C E HC2H3O2 + OH- 0.115 -x 0.115-x 0 x x 1  10 14 Kb for C2H3O2- = = 5.6x10-10 1.8 x10 5 x = [OH-] = 0 x x [ HC 2 H 3O2 ][OH  ] Kb = [C 2 H 3O2 ] 5.6x10-10 x2 = 0.115 0.115 5.6  10 10  = 8.0x10-6 pOH = -log 8.0x10-6 = 5.10 pH = 14 – 5.10 = 8.90  At the equivalence point for a WA/SB titration, the pH > 7 due to the OH- produced by the conjugate base hydrolysis reaction. c) 180.0 mL of NaOH are added from part b, moles HC2H3O2 = 2.50x10-2 moles HC2H3O2  1 L  0.150 moles NaOH   moles NaOH = 180.00 ml    2.70x10-2 moles NaOH 1000 mL L    moles excess base = 2.70x10-2 moles - 2.50x10-2 moles = 2.0x10-3 moles NaOH M OH- = M NaOH = pOH = -log 8.7x10-3 = 2.06 2.0 x10 3 moles  0.2300 L 8.7x10-3 M OH- pH = 14 – 2.06 = 11.94 *Excess NaOH remains - this is the primary source of OH-. We can neglect the hydrolysis of the conjugate base because this would contribute a relatively small amount of OH- compared to the amount that comes directly from the excess NaOH. 19. How many milliliters of 0.95M sodium hydroxide must be added to 35.0 mL of 0.85M acetic acid to reach the equivalence point? Given: Ka for acetic acid is 1.8 x 10-5 A) What is the pH before any base is added? __2.41____ (weak acid, ICE table) B) What is the pH at the equivalence point? ___9.20___ (conjugate base of acid, use Kb in ICE table) Page 8 of 15 C) What is the pH when 15.00 mL of base has been added? __4.71___ (buffer zone) D) What is the pH when 40.00 mL of base has been added? __13.04___ (use excess base to find pH) A) Before base is added, this is a weak acid problem. Set up ICE table and use Ka of acid: Ka = 1.8 x 10-5 HA(aq) + H2O(l)  H3O+(aq) A-(aq) Initial 0.85 M 0 0 Change -x +x +x Equilibrium 0.85 – x X x 2 -5 2 Ka = x / (0.85 – x) = 1.8 x 10 assume x is small: x / 0.85 = 1.8 x 10-5 x = 3.912 x 10-3 M = [H3O+] -3 -3 Check x: (3.912 x 10 / 0.85) x 100% = 4.602 x 10 (Yeah!) B) Step 3 of titration (at the equivalence point). Find the volume of NaOH by stoichiometry: 0.0350L (0.85 mol/L)( 1 NaOH / 1 acid)(1000mL / 0.95M) = 31.316 mL = Vb = 31.31 mL First they react together one way since NaOH is strong. Set up an initial final table. Calculate moles of each. Note they are equal since we are at the equivalence point cause nothing is in excess at the equivalence point, only product salt exists in the beaker. NaOH + CH3COOH  H2O (l) + NaCH3COO 0.02975 moles 0.02975 moles --- 0 all reacts --- +0.02975 all reacts 0 0 --0.02975 moles Now what happens? No acid left, no base left = equivalence point!!! We have only product. But this salt is not neutral - it contains the conjugate base acetate ion. Basic ions react in water just like any base. We need the molarity of acetate ion. Note the new volume of 66.316 mL. The basic salt NaCH3COO will dissolve completely leaving 0.02975 moles sodium ion and 0.02975 moles acetate ion. Acetate ion is basic and will react further. Sodium ions are neutral and will not react further. We must put concentrations in ICE tables, so we need the molarity of the acetate ion. M CH3COO- is 0.02975 moles / 0.066316 L = 0.4486 M Set up an ICE table for the C. base reacting with water. H2O (l) + CH3COO OH+ CH3COOH --- 0.4486 M 0 0 --- -x +x +x --0.4486 - x x x -10 This is a base reaction, need Kb. Get it from Kw / Ka. Kb = 5.556 x 10 = x2 / 0.4486 x = 1.5787 x 10-5M (note I’m not rounding anything till the final answer) pOH = 4.80 so pH = 9.20 (two decimal places since the M given have two sig figs) C) This is in the buffer zone. Calculate the concentration of acid and conjugate base to use HendersonHasselbalch equation. Page 9 of 15 HA: 0.85M * 35.00 mL = 29.75 mmol OH-: 0.95 M * 15.00 mL = 14.25 mmol Mmol acid in excess: 29.75 mmol – 14.25 mmol = 15.50 mmol / total volume (50.00 mL) = 0.31 M Mmol base (from OH-): 14.25 mmol / total volume (50.00 mL) = 0.285 M pH = pKa + log ([A-] / [HA]) = 4.7447 + log (0.285 / 0.31) = 4.71 D) Excess base determines pH here. 0.95 M * 40.00 mL base = 38 mmol – 29.75 mmol HA 8.25 mmol base / total volume (75.00 mL) = 0.11 M = [OH-] pOH = -log(0.11) = 0.959, pH = 14 – 0.959 = 13.04 20. How many milliliters of 0.35M sodium hydroxide must be added to 25.0 mL of 0.45M acetic acid to reach the equivalence point? What is the pH at the equivalence point? Given: Ka for acetic acid is 1.8 x 10-5 0.0250L (0.45 mol/L)( 1 NaOH / 1 acid)(1000mL / 0.35M) = Vb = 32 mL First they react together one way since NaOH is strong. Set up an initial final table. Calculate moles of each. Note they are equal since we are at the equivalence point. NaOH + CH3COOH  H2O (l) + NaCH3COO 0.0113 moles 0.0113 moles --- 0 all reacts all reacts --- +0.0113 0 0 --- 0.0113 moles Now what happens? No acid left, no base left = equivalence point!!! We have only product. But this salt is not neutral - it contains a C. base acetate ion. Bases react in water. We need the molarity of acetate ion. Note the new volume of 57.0 mL. NaCH3COO will dissolve completely leaving 0.0113 moles sodium ion and 0.0113 moles acetate ion. Acetate ion is basic and will react further. Sodium ions are neutral and will not react further. M CH3COO- is 0.0113 moles / 0.0570 L = 0.198 M Set up an ICE table for the C. base reacting with water. H2O (l) + CH3COO-  OH- + CH3COOH --- 0.198 M 0 0 --- -x +x +x --0.198 - x x x This is a base reaction, need Kb. Get it from Kw / Ka. Kb = 5.56 x 10-10 = x2 / 0.198 x = 1.05 x 10-5M pOH = 4.98 Page 10 of 15 pH = 9.02 (two decimal places since the M given have two sig dig, I just don't round until the end) Solubility Equilibria, Ksp 21. Solubility product constants are usually specified for 250 C. Why does the Ksp value for a chemical compound depend on the temperature? Ksp depends on temperature because solubility depends on temperature. Generally, solids become more soluble as the temperature of the solution increases. As a result, Ksp values of solids tend to increase as the temperature increases. 22. Draw a representation of a solution past saturation of calcium phosphate. Formula = ___ Ca3(PO4)2__ There should be solid Ca3(PO4)2 on the bottom of the beaker and then calcium ions and phosphate ions in solution in the correct ration: 3 Ca2+ for every 2 PO43- ions 23. The Ksp for nickel (II) hydroxide is 5.47 x 10-16. What is the base dissociation constant for nickel (II) hydroxide? 5.47 x 10-16. Because nickel (II) hydroxide dissociates to become a base, the Ksp and Kb values are identical. 24. What is the concentration of a saturated silver acetate solution? Ksp(AgC2H3O2) = 1.94 x 10-3. 25. Since Ksp = [Ag+][C2H3O2-], and the concentration of silver ions is the same as the concentration of acetate ions, we can set up the following equation: 1.94 x 10-3 = x2 x = 0.0440 M What is the concentration of a saturated lead chloride solution? Ksp(PbCl2) = 1.17 x 10-5. Ksp = [Pb+2][Cl-]2. Since the concentration of chloride ions is twice that of lead (II) ions, this boils down to the following equation: 1.17 x 10-5 = (x)(2x)2 1.17 x 10-5 = 4x3 x = 0.0143 M 26. I have discovered a new chemical compound with the formula A2B. If a saturated solution of A2B has a concentration of 4.35 x 10-4 M, what is the solubility product constant for A2B? Ksp = [A+]2[B2-]. Since the concentration of A is twice that of B, and the concentration of B is 4.35 x 10-4 M, we can set up the following equation: Ksp = [2(4.35 x 10-4 M)]2 [4.35 x 10-4 M] Ksp = 3.29 x 10-10 27. Calculate the solubility of AuCl3(s) in pure water. Ksp for AuCl3 = 3.2  10-25. Page 11 of 15 AuCl3(s)  Au3+ + 3ClI ---0 0 C -x +x +3x E ----x 3x Ksp = [Au3+][Cl-]3 = (x)(3x)3 3.2 x 10-25 = 27x4 x4 = 1.185 x 10-26 x = 3.299 x 10-7 M = molar solubility of AuCl3 in water 3.299 x 10-7 mol/L (303.32 g/mol) = 1.0 x 10-4 g/L 28. What is the solubility for zinc sulfide (ZnS) if the Ksp is 2.1 x 10-25? ZnS(s) Zn2+(aq) + -25 2.1 x 10 = Ksp = [Zn2+][S2-] = x2 x = 4.6 x 10-13 mol / L S-2(aq) gram solubility (not asked for): 4.58x10 -13 mol/L (97.44 g / mol) = 4.5 x 10-11 g/L 29. At 25 °C, 0.0349 g of Ag2CO3 dissolves in 1.0 L of solution. Calculate Ksp for this salt. solubility = 1 mol Ag 2 CO3 0.0349 g Ag 2 CO3 x = 1.3x10-4 M Ag2CO3 275.8 g Ag 2 CO3 1.0 L Ag2CO3(s) 2Ag+(aq) + CO 32  (aq) I C E 0 2x 2x Ksp = [Ag+]2[CO 32  ] 0 x x x = molar solubility of Ag2CO3 = 1.3x10-4 M [CO 32  ] = x = 1.3x10-4 M [Ag+] = 2x = 2(1.3x10-4 M) = 2.6x10-4 M Ksp = [2.6x10-4 ]2[1.3x10-4] = 8.8x10-12 30. Silver phosphate, Ag3PO4, is an insoluble salt that has a Ksp = 1.3 x 10-20. a) Calculate the molar solubility of Ag3PO4 in pure water. Ag3PO4(s) I C E 3Ag+(aq) + PO43-(aq) 0 3x 3x Ksp = [Ag+]3[PO43-] 0 x x Ksp = (3x)3x 1.3x10-20 = 27x4 x4 = 4.8x10-22 x = 4.7x10-6 M = molar solubility of Ag3PO4 in pure water b) Calculate the molar solubility of Ag3PO4 in a solution containing 0.020 M Na3PO4 (a soluble salt). soluble salt: Na3PO4  3Na+ + PO43Phosphate is the common ion: Page 12 of 15 [PO43-] = [Na3PO4] = 0.020 M (since 1 mol Na3PO4 forms 1 mol PO43- ions) Ag3PO4(s) 3Ag+(aq) + PO43-(aq) I C E 0 3x 3x 0.020 x 0.020+x Ksp = [Ag+]3[PO43-] 1.3x10-20 = = (3x)30.020 6.5x10-19 = 27x3 x3 = 2.4x10-20 x = 2.9x10-7M = molar solubility of Ag3PO4 with a common ion  Adding common ion decreases the solubility of Ag3PO4 31. Will the amount of dissolved silver iodide (largely insoluble) increase, decrease, or remain the same if silver nitrate (soluble) is added to a saturated solution of silver iodide? Explain and support with appropriate chemical reactions. AgI(s)  Ag+(aq) + I-(aq) When AgNO3 added it is soluble so add dissolves. Adding silver ions shifts the equilibrium rxn left toward solid AgI. So the solubility and the amount of dissolved AgI decreases. 32. Will the solubility of barium carbonate increase, decrease, or remain the same if solid barium nitrate is added to a saturated solution of barium carbonate? Explain and support with appropriate chemical reactions. BaCO3(s)  Ba+2(aq) + CO32-(aq) When Ba(NO3)2 added it is soluble so it completely dissolves. Adding barium ions shifts the equilibrium rxn left toward the solid. So the solubility and the amount of dissolved BaCO3 decreases. 33. Does AgCl precipitate from a solution containing 1.0 x 10-5 M Cl- and 1.5 x 10-4 M Ag+? Ksp = 1.8 x 10-10 Calculate Q for AgCl(s) Ag+ + ClQ = [Ag+][Cl-] Q = [1.5x10-4][1.0x10-5] = 1.5x10-9 1.5x10-9 > 1.8x10-10; Q > Ksp Equilibrium shifts left & solid forms; AgCl precipitates 34. If you mix 10.0 ml of 0.0010 M Pb(NO3)2 with 5.0 ml of 0.015 M HCl, does PbCl2 precipitate? Ksp of PbCl2 = 1.6 x 10-5 Pb(NO3)2(aq) + 2HCl(aq)  PbCl2(s) + 2HNO3(aq) Net ionic: Pb2+ + 2Cl-  PbCl2(s) Solubility reaction: PbCl2(s) Pb2+ + 2ClPage 13 of 15 Calculate Q for PbCl2: Q = [Pb2+][Cl-]2   10.0 ml  = 6.7x10-4 M Pb2+ [Pb2+] = 0.0010 M Pb2+   10.0 ml  5.0 ml    5.0 ml  = 5.0x10-3 M Cl[Cl-] = 0.015 M Cl-  5 . 0 ml  10 . 0 ml   Q = (6.7x10-4)(5.0x10-3)2 = 1.7x10-8 Q < Ksp, so PbCl2 does not precipitate. 35. If you mix 225.0 mL of 0.015 M aqueous lead(II) nitrate with 125.0 mL of 0.045 M aluminum bromide, does a precipitate form? Ksp for PbBr2 = 6.9 x 10-6. Must show your work mathematically by calculating - no guessing. Precipitation reaction: 3 Pb(NO3)2(aq) + 2 AlBr3(aq)  3 PbBr2(s) + 2 Al(NO3)2(aq) We do not know if enough PbBr2 was made to pass the saturation point, so we will calculate Q based on the Pb and Br ion concentrations. Since soluble the [Pb2+] = .015M Pb(NO3)2 (1 Pb2+ / 1 Pb(NO3)2) = 0.015 M Pb2+ originally Since soluble [Br-] = 0.045 mol AlBr3 3 mol Br  = 0.135 M Br- originally  1L 1 mol AlBr3 Now these two solutions were added together, thus diluted, with a final volume of 350.0 mL  0.015 M Pb 2  225.0 ml   = 9.643 x 10-3 M Pb2+ 350 . 0 ml   In the final mixture: [Pb2+] =   0.135 M Br   125.0 ml   = 4.821 x 10-2 M Br350 . 0 ml   In the final mixture: [Br-] =  Solubility reaction: PbBr2(s)  Pb2+ + 2BrQ = [Pb2+][Br-]2 = (9.643x10-3)(4.821x10-2)2 Q = 2.2x10-5 Q > Ksp, so PbBr2 does precipitate 36. Solid calcium fluoride is added to 1.00 liter of pure water. After several hours of stirring, some of the solid remains undissolved. If the concentration of the calcium ions is 7.2 x 10-5 M, calculate the solubility product (Ksp). CaF2(s)  I -C -x E --- Ca2+(aq) + 2 F-(aq) 0 0 x 2x x 2x Ksp =[Ca2+][F-]2 = x (2x)2 = 4x3 x = 7.2 x 10-5 So substitute and Ksp = 1.5 x 10-12 37. In lab Sally adds 0.0244 grams of solid calcium fluoride to 1.50 liters of pure water and stirs vigorously. For calcium fluoride Ksp = 4.0 x 10-11 . Will the resulting solution be unsaturated, just saturated, or will solid be present? Show your work for full credit. Page 14 of 15 0.0244 g CaF2 ( mol / 78.08 g) = 3.125 x 10-4 mol; divide by 1.50L = 2.083 x 10-4M CaF2 if it all actually dissolves. If all of the solid dissolves there will be [Ca2+] = 2.083 x 10-4M and twice that for [F-] = 4.167 x 10-4M CaF2(s)  Ca2+(aq) + 2 F-(aq) So Q = [Ca2+][F-]2 = 3.62 x 10-11 which is less than Ksp so unsaturated Page 15 of 15