The common ion effect is a part of Le Chatelier s principle.

Transcription

1 Chapter-17 Lesson-1 I Acid-Base Solutions and the Common Ion Effect Adding a salt (an ionic solid) with a common ion to a weak acid solution or a weak base solution will change the equilibrium of the solution and thus will change the ph of the solution. This change is known as the common ion effect. The common ion effect is a part of Le Chatelier s principle. Q1: A 1.00 M solution of HNO 2 has a ph of 1.67 [K a = ]. Determine if the ph will increase, decrease, or remain the same when the following substances are added to the 1.00 M HNO 2. (i) HCl (g) (ii) KOH (s) (iii) KCl (s) (iv) NaNO 2(s) A1: HNO 2 + H 2 O H 3 O 1+ + NO 2 1- (i) Adding the strong acid, HCl, will increase the concentration of H 3 O 1+ forcing the equilibrium of the HNO 2 system to the left. Thus the ph will decrease. (ii) Adding the strong base, KOH, will increase the concentration of OH 1- in the solution. The OH 1- ions will react with and thus remove the H 3 O 1+ ions from the HNO 2 system forcing the equilibrium of the HNO 2 system to the right. Thus the ph will increase. (iii) Adding the neutral salt, KCl, will have no effect on the equilibrium of the HNO 2 system. Thus the ph will remain the same (iv) Adding the basic salt, NaNO 2, will produce free NO 2 1- ions in the solution. This will force the equilibrium of the HNO 2 system to the left, thus the ph will increase.

2 Q2: Lactic acid is HC 3 H 5 O 3 [K a = ]. Will the initial ph of 2.19 for a M solution of lactic acid increase, decrease, or remain the same when solid NaC 3 H 5 O 3(s) is added? A2: HC 3 H 5 O 3(aq) + H 2 O (l) H 3 O 1+ (aq) + C 3 H 5 O 3 1- (aq) Adding solid NaC 3 H 5 O 3(s) to the lactic acid solution will increase the concentration of C 3 H 5 O 3 1- (aq) and push equilibrium to the left. Thus the concentration of H 3 O 1+ will decrease and ph will rise. Q3: Determine the new ph of a 1.00L solution of M HC 3 H 5 O 3 [K a = ] after 20.0 g of NaC 3 H 5 O 3 is added. Assume that the volume remains constant after adding the solid. A3: HC 3 H 5 O 3(aq) + H 2 O (l) H 3 O 1+ (aq) + C 3 H 5 O 3 1- (aq) 20.0 g NaC 3 H 5 O 3 1mol = 0.179mol NaC 3 H 5 O L = M 112 g Assume that you initially have only HC 3 H 5 O 3 and NaC 3 H 5 O 3. HC 3 H 5 O 3(aq) + H 2 O (l) H 3 O (aq) + C 3 H 5 O 3 (aq) I C - X + X + X E X X X K a = [H 3 O 1+ ][ C 3 H 5 O 3 1- ] [HC 3 H 5 O 3 ] = [X][0.179+X] [0.300-X] Since Ka is small, X might be small too in relation to both 0.300M and 0.179M = [X][0.179] [0.300] X = M H 1+ ph = 3.64 X is 0.077% ionization of M HC3H5O3 and 0.13% ionization of M C3H5O3 1-

3 Q4: Washing soda is Na 2 CO 3 [K b = ]. Will the initial ph of for a M solution of washing soda increase, decrease, or remain the same when some solid NaHCO 3 is added into it? A4: CO 3 2- (aq) + H 2 O (l) OH 1- (aq) + HCO 3 1- (aq) Adding solid NaHCO 3 to the Na 2 CO 3 solution will increase the concentration of HCO 3 1- (aq) and push equilibrium to the left. Thus the concentration of OH 1- will decrease and ph will fall. Q5: Determine the new ph of a 1.00L solution of M Na 2 CO 3 [K b = ] after 21.0 g of solid NaHCO 3 is added. Assume that the volume remains constant after adding NaHCO 3. A5: CO 3 2- (aq) + H 2 O (l) OH 1- (aq) + HCO 3 1- (aq) 21.0g NaHCO 3 1 mol = mol HCO L = M 84.0 g Assume that you initially have only CO 3 2- and HCO CO 3 + H 2 O OH HCO 3 I C - X + X + X E X X X **************************************************************************** K a = [OH 1- ][HCO3 1- ] [CO 3 2- ] = [X][0.250+X] [0.100-X] = [X][0.250] [0.100] Since Ka is small, X might be small too in relation to both 0.100M and 0.250M X is 0.072% ionization of M CO3 2- and 0.029% ionization of M HCO3 1- X = M OH 1- poh = 4.14 ph = 9.86

4 Q6: Determine the ph of a 10.0 ml solution of M CH 3 NH 2 [K b = ] mixed with 15.0 ml of M CH 3 NH 3 Cl. Assume that the liquid volumes are additive. A6: CH 3 NH 2 (aq) + H 2 O (l) CH 3 NH 3 1+ (aq) + OH 1- (aq) When mixing two solutions, it is necessary to recalculate the molarities because each solution dilutes the concentration of the other. Remember M c V c = M d V d. for CH 3 NH 2 M c V c = M d V d (0.450 M)(15.0 ml) =(X)(25.0 ml) X = M CH 3 NH 2 for CH 3 NH 3 Cl M c V c = M d V d (0.550 M)(10.0 ml) =(X)(25.0 ml) X = M CH 3 NH 3 Cl Assume that you initially have only CH 3 NH 2 and CH 3 NH 3 Cl. Rx CH 3 NH 2 (aq) + H 2 O (l) 1+ CH 3 NH 3 (aq) + OH 1- (aq) I C -X +X +X E X X X K b = [CH3NH3 1+ ][OH 1- ] [CH 3 NH 2 ] = [0.270+X][X] [0.220-X] = [0.270][X] [0.220] Since Kb is small, X might be small too in relation to both 0.270M and 0.220M X is 0.13% ionization of M CH3NH3 1+ and 0.16% ionization of M CH3NH2 X = M OH 1- poh = 3.44 ph = 10.56

5 Lesson-2 II Buffers Chapter-17 A system whose ph changes only slightly when strong acid or strong base is added is called a buffer system or simply a buffer. A buffer ordinarily contains roughly equal amounts of a weak acid and its conjugate base (as a basic salt). [HA] [MA] To create the buffer system: HC 2 H 3 O 2(aq) H 1+ (aq) + C 2 H 3 O 2 1- (aq) dissolve equal amounts of HC 2 H 3 O 2 and NaC 2 H 3 O 2 in water. A) [H + ] in Solutions Made by Adding HA and A 1- to H2O For the buffer system made by adding equal amounts of HA and NaA to water HA (aq) H 1+ (aq) + A 1- (aq) K a = [H 1+ ][A 1- ] [HA] and [H 1+ ] = Ka[HA] [A 1- ] and ph = pk a + log [A 1- ] [HA] Q1: Calculate the [H + ] and ph at equilibrium of a buffer prepared with mol HC 2 H 3 O 2 (K a = ) and mol NaC 2 H 3 O 2 in 1.00 L water. A1: ph = pk a + log ([A 1- ]/[HA]) ph = -log ( ) + log([0.200]/[0.200]) ph = 4.74 [H + ] = M Since HA is a weak acid assume that [HA] added [HA] at equilibrium. The Henderson-Hasselbalch Equation Note: ph = pka when [HA] = [A 1- ]

6 Q2: Calculate the ph at equilibrium of a buffer system prepared by adding 1.00 g NaHCO 3 and 1.00 g Na 2 CO 3 to1.00 L water. (MM NaHCO 3 = 84.0 g/mol) (MM Na 2 CO 3 = 106 g/mol) (K a HCO 3 1- = ) A2: NaHCO 3 : 1.00 g 1 mole = mol 1.00 L = M 84.0 g Na 2 CO 3 : 1.00 g 1 mole = mol 1.00 L = M 106 g ph = pk a + log ([A 1- ]/[HA]) ph = -log ( ) + log([ ]/[0.0119]) ph = B) Choosing a Buffer System NOTE: If HA and A 1- are present in roughly equal amounts, the ph of the buffer is roughly equal to the pk a of the weak acid. To prepare a buffer of a particular ph, choose a system where the pk a of the weak acid is roughly equal to the desired ph. A buffer has a useable ph range of about ± 1 ph unit of pk a. The buffer capacity (the ability to resist change) is directly related to the initial concentrations of HA and A 1-. Q3: Which of the following conjugate acid-base pairs will not function as a buffer? HF & F 1-, HClO & ClO 1-, HNO 3 & NO 3 1- A3: HNO 3 & NO 3 1- because HNO 3 is a strong acid. Q4: What is the optimal ph buffered by a solution of HCHO 2 and NaCHO 2? (K a of HCHO 2 = ) A4: ph = 3.74 because the pk a of HCHO 2 = Q5: Which of the following 1.0 M conjugate acid-base pairs will best function as a ph 4.00 buffer? (consult the Ka values of the weak acids) HF & NaF, HClO 2 & NaClO 2, or HC 6 H 7 O 6 & NaC 6 H 7 O 6 A5: HC 6 H 7 O 6 & NaC 6 H 7 O 6 because pk a of HC 6 H 7 O 6 = 4.10

7 Q6: How would you prepare a 5.0 L buffer (ph = 4.25) with solid C 6 H 5 COOH (K a = ) and a M C 6 H 5 COONa solution. A6: ph = pk a + log ([A 1- ]/[HA]) 4.25 = -log( ) + log([0.050]/[x]) 4.25 = (-1.30) - (logx) 4.25 = logx log X = X = M 5.0 L 122 g = 27 g C 6 H 5 COOH 1 mol Dissolve 27 g C 6 H 5 COOH in 4.0 L of M C 6 H 5 COONa solution and then bring the buffer solution up to 5.0 L with additional M C 6 H 5 COONa solution. C) Effect of Stressing a Buffer System A buffer resists ph changes because it contains one species (HA) that reacts with OH 1- ions and another species (A 1- ) that reacts with H 1+ ions. HA (aq) + addedoh 1- (aq) H 2 O ( ) + A 1- (aq) A 1- (aq) + addedh 1+ (aq) HA (aq) Thus, added H 1+ or OH 1- is consumed and so ph remains stable. Q7: Calculate the ph of a (0.200 mol HC 2 H 3 O 2 /0.200 mol C 2 H 3 O 2 1- ) buffer (K a = ) after the addition of mol NaOH. A7: HC 2 H 3 O 2(aq) + OH 1- (aq) H 2 O ( ) + C 2 H 3 O 2 1- (aq) HC 2 H 3 O 2 (mol) 1- C 2 H 3 O 2 (mol) Initial Change Equilibrium ph = pk a + log ([C 2 H 3 O 2 1- ]/[HC 2 H 3 O 2 ]) ph = log([0.220]/[0.180]) ph = 4.83 ph of 0.09

8 What happens when the concentration of the buffer system in the previous question is doubled and is then stressed in the same way? Q8: Calculate the ph of a (0.400 mol HC 2 H 3 O 2 /0.400 mol C 2 H 3 O 2 1- ) buffer (K a = ) after the addition of mol NaOH. A8: HC 2 H 3 O 2(aq) + H 2 O ( ) C 2 H 3 O 2 1- (aq) + H 3 O 1+ (aq) HC 2 H 3 O 2 (mol) 1- C 2 H 3 O 2 (mol) Initial Change Equilibrium ph = pk a + log ([C 2 H 3 O 2 1- ]/[HC 2 H 3 O 2 ]) ph = log([0.420]/[0.380]) ph = 4.78 ph of 0.04 Notice that by increasing the concentration of the buffer system the buffer capacity is increased. The ph increased by only Buffers can be made from weak bases and their conjugates. poh = pk b + log [HB 1+ ] [B] Q9: Calculate the final ph of the following ml buffer system (1.0 M NH 3 /1.0 M NH 4 Cl) after the addition ml of 0.10 M HNO 3. (K b of NH 3 = ) A9: NH 3(aq) + H 2 O (l) NH 4 1+ (aq) + OH 1- (aq) ml L 1.0 M = 0.25 mol NH 3 and NH ml L 0.10 M = mol HNO 3 (H 3 O + ) NH 3 (mol) 1+ NH 4 (mol) Initial Change Equilibrium poh = pk b + log ([NH 4 1+ ]/[ NH 3 ]) Also the Henderson-Hasselbalch Equation poh = log([0.26]/[0.24]) poh = 4.77 ph = 9.23

9 Chapter-17 Lesson-3 III Solubility Equilibria of Ionic Solids When ionic solids dissolve in pure water, the extent to which solution occurs depends on a balance between three forces 1) the force of attraction between the water molecules and the ions of the ionic solid (i.e. the ion-dipole forces) 2) the force of attraction between the cations and anions of the ionic solid (i.e. the ionic bonds). 3) the force of attraction between the water molecules themselves (i.e. the hydrogen bonds). If the ion-dipole forces dominate, then we would expect the ionic solid to be very soluble in water. If the ionic bonds dominate, then we would expect the ionic solid to be very insoluble in water. By learning the basic rules of solubility, one can predict whether or not an ionic solid will significantly dissolve in pure water. KCl is soluble in pure water. [34 g in 100 g H 2 O at 25 o C] PbCl 2 is insoluble in pure water. [0.45 g in 100 g H 2 O at 25 o C] AgCl is insoluble in pure water. [ g in 100 g H 2 O at 25 o C] Even the most insoluble ionic solutes dissolve in water - to some extent - before saturation occurs. All saturated solutions of ionic solids are equilibrium systems. dissolving undissolved solid dissolved solid precipitating The concentrations of the ions at equilibrium are governed by an equilibrium constant called the solubility product constant (K sp ). BaCrO 4(s) Ba 2+ (aq) + CrO 4 2- (aq) K sp = [Ba 2+ ][CrO 4 2- ] =

10 A) Solubility of Ionic Solids in Pure Water Given a substance's K sp, (appendix D) it is possible to calculate its solubility (the maximum that will dissolve) in pure water! Q1: Calculate the solubility in water (in mol/l) of Li 2 CO 3 at 25.0 o C. [K sp = ] [Solubility is the maximum that will dissolve.] A1: Li 2 CO 3(s) 2 Li 1+ (aq) + CO 3 2- (aq) let X = the maximum that dissolves Li 2 CO 3(s) 2 Li (aq) + CO 3 (aq) I X C -X +2X +X E 0 2X X K sp = [Li 1+ ] 2 [CO 3 2- ] = [2X] 2 [X] = 4X 3 X = M the maximum that dissolves is M Li 2 CO 3 Q2: Calculate the solubility in water (in g/100 g H 2 O) of Ca(OH) 2 at 25.0 o C. [K sp = ] (mole mass of Ca(OH) 2 = 74g/mol) A2: Ca(OH) 2(s) Ca 2+ (aq) + 2 OH 1- (aq) let X = the maximum that dissolves Ca(OH) 2(s) Ca 2+ (aq) + 2 OH 1- (aq) I X C -X +X +2X E 0 X 2X K sp = [Ca 2+ ][OH 1- ] = [X][2X] 2 = 4X 3 X = M the maximum that dissolves is M Ca(OH) mol/l 74g/mol 1L/1000 g 100g g/100gh 2 O an approximation

11 Given a substance's solubility (the maximum that will dissolve) in pure water, it is possible to calculate its K sp! Q3: A saturated solution of BaF 2 contains 1.38 g/l. Calculate the K sp of BaF 2. (mole mass of BaF 2 = 175g/mol) A3: 1.38 g/l 1mol/175 g = mol/l BaF 2 BaF 2(s) Ba 2+ (aq) + 2 F 1- (aq) BaF 2(s) Ba 2+ (aq) + 2 F 1- (aq) I C E K sp = [Ba 2+ ][F 1- ] 2 K sp = [ ][0.0158] 2 K sp = Q4: If the solubility of solid Ca(OH) 2 is 0.19g/100g H 2 O at 0.0 o C, then calculate its K sp at 0.0 o C. (mole mass of Ca(OH) 2 = 74g/mol) A4: 0.19 g 1 mol/ 74g = mol Ca(OH) 2 in 100 g H 2 O mol/0.100 L(100 g H2O L solution) = M Ca(OH) 2 Ca(OH) 2(s) Ca 2+ (aq) + 2 OH 1- (aq) Ca(OH) 2(s) Ca 2+ (aq) + 2 OH 1- (aq) I C E K sp = [Ca 2+ ][OH 1- ] 2 K sp = [0.026][0.052] 2 K sp = it s more than in question #2!

12 Given a substance's K sp, it is possible to calculate the minimum volume of water necessary dissolve a given mass of substance! Q5: How many milliliters of a saturated PbCl 2 solution will contain 1.4 g of PbCl 2? (K sp = ) A5: If you can find the moles of PbCl 2 that will saturate 1.00 L of water, then you can find the grams of PbCl 2 that will saturate 1.0 L of water, and then you can find the volume of water that would be saturated by 1.4 g of PbCl 2! PbCl 2(s) Pb 2+ (aq) + 2 Cl 1- (aq) let X = the maximum that dissolves PbCl 2(s) Pb 2+ (aq) + 2 Cl 1- (aq) I X C -X +X +2X E 0 X 2X K sp = [Pb 2+ ][Cl 1- ] = [X][2X] 2 = 4X 3 X = M PbCl 2 Thus, mole PbCl 2 will saturate 1.0 L of water mol 278 g = 4.4 g PbCl 2 will saturate 1.0 L of water 1 mol Now set up a proportion 4.4 g PbCl 2 = 1.4 g PbCl L H 2 O X L H 2 O X = 0.32 L H 2 O 320 ml H 2 O

13 Q6: Calculate the solubility in water (in mol/l) of Ag 2 SO 4 at 25.0 o C. [K sp = ] [Solubility is the maximum that will dissolve.] A6: Ag 2 SO 4(s) 2 Ag 1+ (aq) + SO 4 2- (aq) let X = the maximum that dissolves Ag 2 SO 4(s) 2 Ag 1+ (aq) + I X C -X +2X +X E 0 2X X K sp = [Ag 1+ ] 2 [SO 4 2- ] = [2X] 2 [X] = 4X 3 SO 4 2- (aq) X = M the maximum that dissolves is M Ag 2 SO 4 Q7: If the solubility of solid PbI 2 is 0.069g/100g H 2 O at 25 o C, then calculate its K sp at 25 o C. (mole mass of PbI 2 = 461g/mol) A7: g 1 mol/461g = mol PbI mol/0.100 L(100 g H2O L solution) = M PbI 2 PbI 2 (s) Pb 2+ (aq) + 2 I 1- (aq) PbI 2 (s) Pb 2+ (aq) + 2 I 1- (aq) I C E K sp = [Pb 2+ ][ I 1- ] 2 K sp = [ ][ ] 2 K sp =

14 Q8: What volume in milliliters of a saturated CaCrO 4 solution will contain 1.0 g of CaCrO 4 at 25 o C? (K sp = ) A8: If you can find the moles of CaCrO 4 that will saturate 1.00 L of water, then you can find the grams of CaCrO 4 that will saturate 1.0 L of water, and then you can find the volume of water that would be saturated by 1.0 g of CaCrO 4! CaCrO 4(s) Ca 2+ (aq) + CrO 4 2- (aq) CaCrO 4 (s) Ca (aq) + CrO 4 (aq)) I X C -X +X +2 E 0 X 2 K sp = [Ca 2+ ][ CrO 4 2- ] = [X][X] = X 2 X = M CaCrO 4 Thus, mole CaCrO 4 will saturate 1.0 L of water mol 156 g = 4.2 g CaCrO 4 will saturate 1.0 L of water 1 mol Now set up a proportion 4.2 g CaCrO 4 = 1.0 g CaCrO L H 2 O X L H 2 O X = 0.24 L H 2 O 240 ml H 2 O

15 Lesson-4 Chapter-17 B) Solubility of Ionic Solids in Salt Solutions Compared with its solubility in pure water, an ionic solid is less soluble in an aqueous solution which already contains a common ion with the ionic solid. This is known as the common ion effect. The common ion effect is a part of Le Chatelier s principle. Q1:Consider the equilibrium: MgCO 3(s) Mg 2+ (aq) + CO 3 2- (aq) + 25 kj What effect does each of the following have on the position of the equilibrium? i) the addition of solid MgCl 2. ii) an increase in the temperature of the solution. iii) the addition of solid NaCl. iv) in increase of pressure on the solution. v) the addition of solid NaOH (be careful a side reaction occurs). A1: i) With an increased concentration of aqueous Mg 2+ ions, the equilibrium will shift to the left (common ion effect) ii) The equilibrium will shift to the left since the forward reaction is exothermic and an increase in temperature will favor the endothermic (reverse) reaction. iii) With no common ions, the equilibrium remains the same! iv) Since there is no gaseous reactants or products in the system, then pressure will have no effect on the system. v) Mg 2+ ions are insoluble with OH 1- ions. Thus the precipitate Mg(OH) 2 will form. This will cause a decrease in aqueous Mg 2+ ions and thus the equilibrium will shift to the right.

16 Q2: The solubility of solid MgCO 3 in pure water is g/l at 25 o C. Would the solubility of solid MgCO 3 be greater, less, or the same in a M solution of MgCl 2? A2: Since you are dissolving solid MgCO 3 in a solution that already contains aqueous magnesium ions, the solubility of solid MgCO 3 would be less. Q3: Determine the solubility (in g/l) of solid MgCO 3 at 25 o C in a M solution of MgCl 2. (K sp of MgCO 3 = ) A3: Since you are looking for MgCO 3 s solubility, then MgCO 3(s) Mg 2+ (aq) + CO 3 2- (aq) is the equilibrium system. Remember, there already is M Mg 2+ in the water to begin with (initially). let X = the maximum that dissolves MgCO 3(s) Mg 2+ (aq) + 2- CO 3 (aq) I X C -X +X +X E X X K sp = [Mg 2+ ][CO 3 2- ] = [X][ X] since Ksp is very small, then +X can be ignored [X][0.0025] X = M [X is only 0.32% of M] X = mol 84.3 g = g/l MgCO 3 L mole

17 Q4: Determine the molar solubility of PbCl 2 (K sp = ) in M CaCl 2. A4: PbCl 2(s) Pb 2+ (aq) + 2 Cl 1- (aq) let X = the maximum that dissolves PbCl 2(s) Pb 2+ (aq) + 2 Cl 1- (aq) I X C -X +X +2X E 0 X X M CaCl2 will produce M Cl 1- K sp = [Pb 2+ ][Cl 1- ] = [X][ X] 2 since Ksp is small, then +X may be ignored [X][0.200] 2 X = M PbCl 2 (X is only 0.22% of M) Q5: Determine the molar solubility of Li 2 CO 3 (K sp = ) in M Na 2 CO 3. A5: Li 2 CO 3(s) 2 Li 1+ (aq) + CO 3 2- (aq) let X = the maximum that dissolves Li 2 CO 3(s) 2 Li (aq) + CO 3 (aq) I X C -X +2X +X E 0 2X X M Na2CO3 will produce M CO3 2- K sp = [Li 1+ ] 2 [CO 3 2- ] = [2X] 2 [0.100+X] since Ksp is small, then +X may be ignored [2X] 2 [0.100] X = M which is HUGE compared to M!!!

18 So by using the method of successive approximations take the huge answer for X (0.206 M) and insert it in for +X = [2X] 2 [0.100+X] the first approximation: = [2X] 2 [ ] X = M the second approximation: = [2X] 2 [ ] X = M the third approximation: = [2X] 2 [ ] X = M the fourth approximation: = [2X] 2 [ ] X = M the fifth approximation: = [2X] 2 [ ] X = M the sixth approximation: = [2X] 2 [ ] With the advent of powerful calculators, students no longer have to use this method of solving math problems. X = M THUS the answer is X = M Li 2 CO 3

19 Q6:Consider the equilibrium: BaCO 3(s) + heat Ba 2+ (aq) + CO 3 2- (aq) What effect does each of the following have on the position of the equilibrium? i) the addition of solid Na 2 CO 3. ii) an increase in the temperature of the solution. iii) the addition of solid NaCl. iv) in increase of pressure on the solution. A6: i) With an increased concentration of aqueous CO 3 2- ions, the equilibrium will shift to the left (common ion effect) ii) The equilibrium will shift to the right since the forward reaction is endothermic and an increase in temperature will favor the endothermic (forward) reaction. iii) With no common ions, the equilibrium remains the same! iv) Since there is no gaseous reactants or products in the system, then pressure will have no effect on the system. Q7: Determine the molar solubility of BaCO 3 (K sp = ) in pure water. A7: BaCO 3 (s) Ba 2+ (aq) + CO 3 2- (aq) BaCO 3 (s) Ba 2+ (aq) + 2- CO 3 (aq) I X C -X +X +X E 0 X X K sp = [Ba 2+ ][ CO 3 2- ] = [X][ X] = [X] 2 X = M BaCO 3

20 Q8: Determine the molar solubility of BaCO 3 (K sp = ) in a M Na 2 CO 3 solution (not pure water). A8: BaCO 3 (s) Ba 2+ (aq) + CO 3 2- (aq) BaCO 3 (s) Ba 2+ (aq) + 2- CO 3 (aq) I X C -X +X +X E 0 X X K sp = [Ba 2+ ][ CO 3 2- ] M Na2CO3 will produce M CO = [X][0.100+X] since Ksp is small, then +X may be ignored [X][0.100] X = M BaCO 3 (X is only % of M) Q9: Determine the solubility (in g/l) of solid BaF 2 at 25 o C in a M solution of BaCl 2. (K sp of BaF 2 = ) A9: BaF 2(s) Ba 2+ (aq) + 2 F 1- (aq) BaF 2(s) Ba 2+ (aq) + 2 F 1- (aq) I X C -X +X +2X E X 2X K sp = [Ba 2+ ][F 1- ] = [0.075+X][2X] [0.075][2X] 2 X = M [X is 3.2% of M] since Ksp is very small, then +2X can be ignored X = mol g = 0.42 g/l BaF 2 L mole

21 Lesson-5 Chapter-17 C) Solubility of Ionic Solids and ph Insoluble ionic compounds are often more soluble if the ph 7. Insoluble hydroxides can readily dissolve in acidic solutions because water is formed which then pulls equilibrium to the right. M(OH) 2(s) + 2 H 3 O 1+ (aq) M 2+ (aq) + 2 H 2 O (l) Insoluble salts with anions of weak acids (ie: F 1-, S 2-, and CO 3 2- ) are readily more soluble in acidic solutions because weak acid molecules are formed which then pulls equilibrium to the right. MX (s) + H 1+ (aq) M 1+ (aq) + HX (aq) Q1: Write the dissolving equation of solid Fe(OH) 2(s) in an acid solution. A1: (i) Fe(OH) 2(s) Fe 2+ (aq) + 2 OH 1- (aq) dissolving in H2O (ii) 2OH 1- (aq) + 2 H 1+ (aq) 2 H 2 O (l) neutralization in acid (iii) Fe(OH) 2(s) + 2 H 1+ (aq) Fe 2+ (aq) + 2 H 2 O (l) net reaction Q2: Write the dissolving equation of solid SrCO 3(s) in an acid soln. A2: (i) SrCO 3(s) Sr 2+ (aq) + CO 3 2- (aq) dissolving in H2O 2- (ii) CO 3 (aq) + 2 H 1+ (aq) H 2 O (l) + CO 2(g) acid equilibrium (iii) SrCO 3(s) + 2 H 1+ (aq) Sr 2+ (aq) + H 2 O (l) + CO 2(g) net reaction Q3: Write the dissolving equation of solid CaF 2(s) in an acid solution. A3: (i) CaF 2(s) Ca 2+ (aq) + 2 F 1- (aq) dissolving in H2O (ii) 2 F 1- (aq) + 2 H 1+ (aq) 2 HF (aq) acid equilibrium (iii) CaF 2(s) + 2 H 1+ (aq) Ca 2+ (aq) + 2 HF (aq) net reaction

22 Q4: Calculate the molar solubility of Fe(OH) 2 (K sp = ) in pure water. A4: Fe(OH) 2(s) Fe 2+ (aq) + 2 OH 1- (aq) Fe(OH) 2(s) Fe 2+ (aq) + 2 OH 1- (aq) I X C -X +X +2X E 0 X 2X K sp = [Fe 2+ ][ OH 1- ] = [X][2X] 2 [X] = M Thus only moles Fe(OH) 2(s) will dissolve in 1.0 L of pure water Q5: Calculate the molar solubility of Fe(OH) 2 (K sp = ) in an acid solution that has been buffered at a ph of A buffered solution keeps the ph fairly constant even when acid or base is added. A5: Fe(OH) 2(s) Fe 2+ (aq) + 2 OH 1- (aq) If ph = 6.20, then poh = 7.80, and [OH 1- ] = M Fe(OH) 2(s) Fe 2+ (aq) + 2 OH 1- (aq) I X C -X +X +2X E 0 X X A ph of 6.20 will produce M OH 1- K sp = [Fe 2+ ][ OH 1- ] = [X][ X] = [X][ ] 2 If the solution is buffered, then the +2X is negligible! [X] = 3.1 M Thus 3.1 moles Fe(OH) 2(s) will dissolve in 1.0 L of the acid solution! This is a lot for a base that is insoluble in pure water!

23 The solubility in solution of any insoluble hydroxide versus ph can be determined as shown by the problem below. Q6: Determine the relationship between the solubility of Fe(OH) 2 (K sp = ) in water and the ph of the water. A6: Fe(OH) 2(s) Fe 2+ (aq) + 2 OH 1- (aq) K sp = [Fe 2+ ][ OH 1- ] 2 = log [Fe 2+ ] + 2 log [ OH 1- ] = log [Fe 2+ ] + 2 (-poh) = log [Fe 2+ ] + 2 (poh) = log [Fe 2+ ] + 2 (14.00-pH) = log [Fe 2+ ] ph = ph = log [Fe 2+ ] ph = log [Fe 2+ At a ph of 6.45, the solubility ] of Fe(OH)2 in solution is 1.0 M. Log [Fe(OH)2] vs ph Log [Fe(OH)2] ph

24 Certain metallic oxides and metallic hydroxides are insoluble in pure water but are soluble in both acidic and basic solutions. These are called amphoteric oxides and bases. Oxides and hydroxide bases of Al 3+, Cr 3+, Zn 2+, and Sn 2+ are amphoteric. Al 2 O 3 & Al(OH) 3 are the most common examples. Example: In pure water, Al 2 O 3(s) forms insoluble Al(OH) 3(s) Al 2 O 3(s) + 3 H 2 O (l) 2 Al(OH) 3(s) In pure water, Al(OH) 3(s) forms insoluble Al(H 2 O) 3 (OH) 3(s). Al(OH) 3(s) + 3 H 2 O (l) Al(H 2 O) 3 (OH) 3(s) In acid, Al(H 2 O) 4 (OH) 2 1+ (aq) forms which is soluble. Al(H 2 O) 3 (OH) 3(s) + H 3 O + (aq) Al(H 2 O) 4 (OH) 2 1+ (aq) + H 2 O (l) In base, Al(H 2 O) 2 (OH) 4 1- (aq) forms which is soluble. Al(H 2 O) 3 (OH) 3(s) + OH - (aq) Al(H 2 O) 2 (OH) 4 1- (aq) + H 2 O (l) Q7: In water, insoluble Cr(OH) 3(s) exists as Cr(H 2 O) 3 (OH) 3(s). Write a reaction that shows Cr(OH) 3(s) dissolving in acid. A7: Cr(H 2 O) 3 (OH) 3(s) + H 3 O 1+ (aq) Cr(H 2 O) 4 (OH) 2 1+ (aq) + H 2 O (l) Q8: In water, insoluble Cr(OH) 3(s) exists as Cr(H 2 O) 3 (OH) 3(s). Write a reaction that shows Cr(OH) 3(s) dissolving in base. A8: Cr(H 2 O) 3 (OH) 3(s) + OH 1- (aq) Cr(H 2 O) 2 (OH) 4 1- (aq) + H 2 O (l)

25 Q9: In water, insoluble Zn(OH) 2(s) exists as Zn(H 2 O) 2 (OH) 2(s). Write a reaction that shows Zn(OH) 2(s) dissolving in acid. A9: Zn(H 2 O) 2 (OH) 2(s) + H 3 O 1+ (aq) Zn(H 2 O) 3 (OH) 1+ (aq) + H 2 O (l) Q10: A saturated solution of Zn(OH) 2 [K sp = ] has a solubility in pure water of g/100 g H 2 O. This solution has a ph of What would be the solubility of Zn(OH) 2 (in g/100 g solution) in an acid solution that is buffered at ph 6.00? A10: Zn(OH) 2(s) Zn 2+ (aq) + 2 OH 1- (aq) If ph = 6.00, then poh = 8.00, and [OH 1- ] = M Zn(OH) 2(s) Zn 2+ (aq) + 2 OH 1- (aq) I X C -X +X +2X E 0 X X A ph of 6.00 will produce M OH 1- K sp = [Zn 2+ ][ OH 1- ] = [X][ X] = [X][ ] 2 If the solution is buffered, then the +2X is negligible! [X] = 0.20 M Thus 0.20 moles Zn(OH) 2(s) will dissolve in 1.0 L ( 1000 g) of the acid solution and in 100 g of acid solution, moles of Zn(OH) 2 would dissolve mol Zn(OH) g = 2.0 g Zn(OH) 2 100g acid solution 1 mole 100g acid solution

26 Lesson-6 Chapter-17 D) Solubility of Ionic Solids that form Complex Ions Many ionic solids are insoluble in pure water. AgCl (s) Ag 1+ (aq) + Cl 1- (aq) K sp = CuCO 3(s) Cu 2+ (aq) + CO 3 2- (aq) K sp = Cr(OH) 3(s) Cr 3+ (aq) + 3 OH 1- (aq) K sp = Certain Lewis bases are able to form stable complex ions with certain metallic cations (the larger the Kf value the more stable the complex ion). Ag 1+ (aq) + 2 CN 1- (aq) Ag(CN) 2 1- (aq) K f = Cu 2+ (aq) + 4 NH 3(aq) Cu(NH 3 ) 4 2+ (aq) K f = Cr 3+ (aq) + 4 OH 1- (aq) Cr(OH) 4 1- (aq) K f = Thus, the presence of certain Lewis bases will greatly increase the solubility of certain insoluble ionic compounds if a stable complex ion is able to form. AgCl (s) + 2 CN 1- (aq) Ag(CN) 2 1- (aq) + Cl 1- (aq) K eq = K sp K f K eq = CuCO 3(s) + 4 NH 3(aq) Cu(NH 3 ) 4 2+ (aq) + CO 3 2- (aq) K eq = K sp K f K eq = 1200 Cr(OH) 3(s) + OH 1- (aq) Cr(OH) 4 1- (aq) K eq = K sp K f K eq = 1.3

27 The presence of certain Lewis bases will greatly decrease the presence of certain free aqueous metallic cations if a stable complex ion is able to form. AgNO 3(s) Ag 1+ (aq) + NO 3 1- (aq) Ag 1+ (aq) + 2 CN 1- (aq) Ag(CN) 2 1- (aq) K f = AgNO 3(s) + 2 CN 1- (aq) Ag(CN) 2 1- (aq) + NO 3 1- (aq) K eq = Q1: Write the equation for the dissolution of solid silver iodide in an aqueous solution of ammonia. A1: AgI (s) + 2 NH 3(aq) Ag(NH 3 ) 2 1+ (aq) + I 1- (aq) Q2: Calculate the equilibrium constant for the dissolution of solid silver chloride in a solution of hydrocyanic acid. K sp AgCl = , K a HCN = , & K f Ag(CN) 2 1- = A2: AgCl (s) + 2 HCN (aq) Ag(CN) 2 1- (aq) + Cl 1- (aq) + 2 H 1+ (aq) AgCl (s) Ag 1+ (aq) + Cl 1- (aq) K sp = HCN (aq) H 1+ (aq) + CN 1- (aq) K a = Ag 1+ (aq) + 2 CN 1- (aq) Ag(CN) 2 1- (aq) K f = **************************************************************************** AgCl (s) Ag 1+ (aq) + Cl 1- (aq) K sp = HCN (aq) 2 H 1+ (aq) + 2 CN 1- (aq) (K a = ) 2 + Ag 1+ (aq) + 2 CN 1- (aq) Ag(CN) 2 1- (aq) K f = AgCl (s) + 2 HCN (aq) Ag(CN) 2 1- (aq) + Cl 1- (aq) + 2 H 1+ (aq) K eq = K sp K a 2 K f K eq =

28 Q3: To what final concentration of CN 1- must a solution contain in order to have just dissolved mol of AgCl (s) in 1.0 L of solution? K sp of AgCl = & K f of Ag(CN) 2 1- = Q5: When 0.82 g of ZnCl 2 is dissolved 255 ml of M NaCN, A3: AgCl (s) + 2 CN 1- (aq) Ag(CN) 2 1- (aq) + Cl 1- (aq) K eq = AgCl (s) + 2 CN 1-1- (aq) Ag(CN) 2 (aq) + Cl 1- (aq) I M X C M M M E X M M K eq = [Ag(CN) 2 1- ][Cl 1- ] [CN 1- ] = [0.070 M][ M] [X] 2 X = M CN 1- Q4: Calculate the molar solubility of solid AgI in a 2.5 M solution of NH 3(aq). K sp of AgI = & K f of Ag(NH 3 ) 2 1+ = A4: AgI (s) + 2 NH 3(aq) Ag(NH 3 ) 2 1+ (aq) + I 1- (aq) K eq = AgI (s) NH 3(aq) Ag(NH 3 ) 2 (aq) + I 1- (aq) I X C - X - 2X + X + X E X X X K eq = [Ag(NH 3 ) 2 1+ ][I 1- ] [NH 3 ] = [X][X] [2.5-2X] 2 X = M AgI

29 what are the final concentrations of Zn 2+, CN 1-, Zn(CN) 4 2-, Na 1+, and Cl 1- present in the solution? K f of Zn(CN) 4 2- = A5: Zn 2+ (aq) + 4 CN 1- (aq) Zn(CN) 4 1- (aq) 0.82 g 1 mole 1 = M ZnCl g L Zn 2+ (aq) + 4 CN 1-1- (aq) Zn(CN) 4 (aq) I M M C M M M E M These final concentrations assume that the equilibrium went completely to the right and that ALL the Zn 2+ ions are in the Zn(CN) 1-4. But there must be some free Zn 2+ ions, so the equilibrium of the system shifts back slightly to the left! 1- Zn(CN) 4 (aq) Zn 2+ (aq) + 4 CN 1- (aq) I M M C - X + X + 4X E X X X K eq = [Zn 2+ ] [CN 1- ] 4 [Zn(CN) 4 1- ] = [X] [ X] 4 [0.024-X ] = [X] [0.054] 4 [0.024] X = M Zn 2+ Since Keq is small, X might be small in relation to both and Final Concentrations: [Zn 2+ ] = M [CN 1- ] = M [Zn(CN) 2-4 ] = M [Na 1+ ] = M [Cl 1- ] = M

30 Q6: Write the equation for the dissolution of solid copper (II) carbonate in a solution of cyanide. Each copper (II) ion complexes with four cyanide ions. A6: CuCO 3(s) + 4 CN 1- (aq) Cu(CN) 4 2- (aq) + CO 3 2- (aq) Q7: Calculate the equilibrium constant for the dissolution of solid copper (II) carbonate in an aqueous solution of sodium cyanide. K sp CuCO 3 = & K f Cu(CN) 4 2- = A7: CuCO 3(s) Cu 2+ (aq) + CO 3 2- (aq) K sp = Cu 2+ (aq) + 4 CN 1- (aq) Cu(CN) 4 2- (aq) K f = CuCO 3(s) + 4 CN 1- (aq) Cu(CN) 4 2- (aq) + CO 3 2- (aq) K eq = K sp K f K eq = Q8: To what final concentration of CN 1- must a solution contain in order to have just dissolved 10.0 g ( mol) of CuCO 3(s) in 1.0 L of solution? K sp CuCO 3 = & K f Cu(CN) 4 2- = A8: CuCO 3(s) + 4 CN 1- (aq) Cu(CN) 4 2- (aq) + CO 3 2- (aq) K eq = CuCO 3(s) + 4 CN (aq) Cu(CN) 4 (aq) + CO 3 (aq) I M X M M M C M M M M E M X M M K eq = [Cu(CN) 4 2- ][ CO 3 2- ] [CN 1- ] = [ M][ M] [X] 4 X = M CN 1-

31 Chapter-17 Lesson-7 IV Precipitation Reactions When solutions of two different electrolytes (acid, base, or salt) are mixed, an insoluble solid sometimes forms! This solid is called a precipitate. The chemical reaction by which a precipitate is formed is called precipitation. A precipitation reaction can be written in its molecular form. H 2 SO 4(aq) + Ba(OH) 2(aq) 2 HOH ( ) + BaSO 4(s) A precipitation reaction can be written in its net ionic form. 2 H 1+ (aq) + SO 4 2- (aq) + Ba 2+ (aq) + 2 OH 1- (aq) 2 HOH ( ) + BaSO 4(s) By learning the basic rules of solubility, one can predict whether or not an ionic solid will significantly dissolve in water. Ions that form Soluble Compounds Group-1 cations 1+ & NH 4 NO 3 1- C 2 H 3 O 2 1- (CH 3 COO 1- ) 1- HCO 3 1- ClO 3 1- ClO 4 Cl 1-, Br 1-, I 1- SO 4 2- Exceptions Ag 1+, Pb 2+, Hg 2 2+ Ag 1+, Pb 2+, Ca 2+, Sr 2+, Ba 2+ Ions that form Insoluble Compounds CO 3 2- CrO 4 2- PO 4 3- S 2- OH 1- Exceptions Group-1 cations 1+ & NH 4 Group-1 cations & NH 1+ 4, & Mg 2+, Ca 2+ Group-1 cations 1+ & NH 4 Group-1 cations 1+ & NH 4 Group-1 cations & NH 1+ 4, & Sr 2+, Ba 2+. Ca 2+ is slightly soluble.

32 Q1: Write (i) the molecular equation and (ii) the net ionic equation for the precipitation reaction that occurs between aqueous solutions of silver nitrate and potassium chromate. A1: (i) 2 AgNO 3(aq) + K 2 CrO 4(aq) 2 KNO 3(aq) + Ag 2 CrO 4(s) (ii) 2 Ag 1+ (aq) + 2 NO 3 1- (aq) + 2 K 1+ (aq) + CrO 4 2- (aq) 2 Ag 1+ (aq) + CrO 4 2- (aq) Ag 2 CrO 4(s) 2 K 1+ (aq) + 2 NO 3 1- (aq) + Ag 2 CrO 4(s) Q2: Write (i) the molecular equation and (ii) the net ionic equation for the precipitation reaction that occurs between aqueous solutions of lithium bromide and magnesium sulfate. A2: (i) 2 LiBr (aq) + MgSO 4(aq) Li 2 SO 4(aq) + MgBr 2(aq) (ii) 2 Li 1+ (aq) + 2 Br 1- (aq) + Mg 2+ (aq) + SO 4 2- (aq) no net ionic reaction occurs! 2 Li 1+ (aq) + SO 4 2- (aq) + Mg 2+ (aq) + 2 Br 1- (aq) IN FACT NO CHEMICAL REACTION OCCURS! Q3: If a mixture containing M Na 1+ ions, M Pb 2+ ions, and M Ag 1+ ions, is treated dropwise with 1.00 M HCl, in what order will the ions precipitate? A3: Ag 1+ ions (K sp AgCl = ) and then Pb 2+ ions (K sp PbCl 2 = ). The Na 1+ ions would probably not precipitate. Q4: How could a mixture of K 1+ ions and Ca 2+ ions be separated? A4: Add dilute sulfuric acid to precipitate the Ca 2+ ions. Then filter the mixture to capture the solid CaSO 4.

33 Q5: ml of M lead (II) nitrate is added to 50.0 ml of M HCl. (i) Write the balanced molecular equation for the reaction. (i) Write the net ionic equation for the reaction. (iii) Determine the limiting reagent. (iv) How many grams of precipitate are formed if there is 100% yield? (v) How many moles of the non-limiting reagent remain dissolved in the final mixture? (vi) Calculate the molar concentration for each of the aqueous ions present in the final mixture (assume volumes are additive). A5: (i) Pb(NO 3 ) 2(aq) + 2 HCl (aq) PbCl 2(s) + 2 HNO 3(aq) M = Mol L (ii) Pb 2+ (aq) + 2 Cl 1- (aq) PbCl 2(s) (iii) mol Pb(NO 3 ) L = mol 1 mol = L mol HCl L = mol 2 mol = L (vi) mol HCl L 1 mol PbCl g = L 2 mol HCl 1 mol PbCl g PbCl 2 (v) mol Pb(NO 3 ) 2 initial mol Pb(NO 3 ) 2 reacted = mol Pb(NO 3 ) 2 left (vi) mol Pb(NO 3 ) mol Pb 2+ = M Pb L mol Pb(NO 3 ) mol NO 3 1- = M NO L mol HCl mol H 1+ = M H L

34 An electrolyte's ion product (Q sp ) is the instantaneous product of the electrolyte's cation and anion concentrations in the solution. Q sp = [cation] X [anion] Y - If Q sp < K sp, then the solution is unsaturated (not at equilibrium) and no precipitate forms. - If Q sp = K sp, then the solution is just saturated (at equilibrium) and is at the point of precipitation. - If Q sp > K sp, then the solution is supersaturated (beyond equilibrium) and precipitate forms until P = K sp. Q6: Will a precipitate form if a solution mixture suddenly contains [Pb 2+ ] = M and [Cl 1- ] = M. (K sp = ) A6: PbCl 2(s) Pb 2+ (aq) + 2 Cl 1- (aq) Q sp = [Pb 2+ ][ Cl 1- ] 2 Q sp = [0.075][0.025] 2 = M Q sp > K sp yes, precipitate forms! Pb 2+ (aq) + 2 Cl 1- (aq) PbCl 2(s) Q7: A student mixes 250 ml of M Pb(NO 3 ) 2 with 250 ml of M HCl. Will a precipitate form? (K sp = ) A7: PbCl 2(s) Pb 2+ (aq) + 2 Cl 1- (aq) Q sp = [Pb 2+ ][Cl 1- ] 2 Be careful! When mixing solutions the volume changes and thus the ion concentrations change as well! [Pb 2+ ] goes from M to M because volume doubles! [ Cl 1- ] goes from M to M because volume doubles! Q sp = [0.030][0.020] 2 = M Q sp < K sp no precipitate forms!

35 Q8: If a mixture containing M F 1- ions, M SO 4 2- ions, and M NO 3 1- ions, is treated dropwise with 1.00 M CaCl 2, in what order will the ions begin precipitate? (K sp CaF 2 = and K sp CaSO 4 = ) A8: F 1- ions will precipitate first and then SO 4 2- ions will precipitate. The NO 3 1- ions would probably not precipitate. Q9: Will a precipitate form if a solution mixture suddenly contains [Ca 2+ ] = M and [F 1- ] = M. (K sp CaF 2 = ) A9: CaF 2(s) Ca 2+ (aq) + 2 F 1- (aq) Q sp = [Ca 2+ ][ F 1- ] 2 Q sp = [ ][ ] 2 = M Q sp < K sp no, a precipitate will not form! Q10: A student mixes 250 ml of M Ca(NO 3 ) 2 with 250 ml of M H 2 SO 4. Will a precipitate form? (K sp CaSO 4 = ) Assume that the volumes are additive. A10: CaSO 4(s) Ca 2+ (aq) + SO 4 2- (aq) Q sp = [Ca 2+ ][SO 2-4 ] [Ca 2+ ] goes from M to M because volume doubles! [SO 2-4 ] goes from M to M because volume doubles! Q sp = [0.0050][0.0050] = M Q sp > K sp yes, a precipitate does form!

36 Chapter-17 Lesson-8 V Acid-Base Titrations A) Acid-Base Indicators Acid-base titration is a laboratory process in which equivalent quantities of acid and base are reacted. The equivalence point occurs when moles of H 1+ = moles of OH 1- In an acid-base titration, an indicator is used to signal the end of the reaction (this is how you know when to stop adding reagent). The end point occurs when the acid-base indicator changes color. The choice of acid-base indicator is determined by the type of titration (reaction) being performed. Type of Reaction strong acid - weak base strong acid - strong base weak acid - strong base ph at Equivalence Point Below 7 Equal to 7 Above 7 Appropriate Acid-Base Indicator Color Change and ph Range of Color Change methyl orange red to orange ( ) bromcresol green yellow to blue ( ) bromthymol blue yellow to blue ( ) litmus red to blue ( ) phenolphthalein colorless to pink (8-9) thymol blue yellow to blue ( ) Acid-base indicators are weak acids whose pk a values are within the ph range of color change. HIn + OH 1- In 1- + HOH Color 1 base Color 2 acid Q1: Predict the ph of a solution if a sample turns bromcresol green blue and bromthymol blue yellow. A1: The ph of the solution is between 5.4 and 6.0.

37 Q2: Predict the ph of a solution if a sample turns bromthymol blue blue and thymol blue yellow. A2: The ph of the solution is between 7.6 and 8.0. B) Acid-Base Titration Problems Q3: How many milliliters of 0.50 M H 2 SO 4 would be needed to titrate 8.0 ml of 0.75 M NH 3? (i) Choose an appropriate indicator for this titration. (ii) How many milliliters of acid are required for neutralization? A3: (i) Methyl orange or bromcresol green (ii) H 2 SO NH 3 (NH 4 ) 2 SO L 0.75 mol NH3 1 mol H2SO4 = mol H 2 SO 4 1 L 2 mol NH 3 M = mol L 0.50M H 2 SO 4 = mol H2SO4 X X = 6.0 ml H 2 SO 4 Q4: It takes 25.0 ml of M HC 2 H 3 O 2 to neutralize 9.0 ml of NaOH. (i) Choose an appropriate indicator for this titration. (ii) What is the molar concentration of the NaOH? A4: (i) Thymol blue or phenolphthalein (ii) HC 2 H 3 O 2 + NaOH HOH + NaC 2 H 3 O L mol HC2H3O2 1 mol NaOH 1 = 1 L 1 mol HC 2 H 3 O L X = M NaOH

38 The titration equation: a M a V a = b M b V b - a equals the number of ionizable H 1+ ions from the acid - Ma equals the molarity of the acid - Va equals the volume of the acid used in the titration - b equals the number of OH 1- ions from the base - Mb equals the molarity of the base - Vb equals the volume of the base used in the titration Q5: It takes 30.0 ml of M NH 3 to neutralize 18.0 ml H 2 SO 4. (i) Choose an appropriate indicator for this titration. (ii) What is the molar concentration of the? A5: (i) Methyl orange or bromcresol green (ii) a M a V a = b M b V b (2)(X)(18.0) = (1)(0.024)(30.0) X = M H 2 SO 4 Q6: It takes 26.2 ml of M HCl to neutralize 20.0 ml Ba(OH) 2. (i) Choose an appropriate indicator for this titration. (ii) What is the molar concentration of the Ba(OH) 2? A6: (i) Bromthymol blue or litmus (ii) a M a V a = b M b V b (1)(0.120)(26.2) = (2)(X)(20.0) X = M Ba(OH) 2 Q7: It takes 15.0 ml of M HCHO 2 to neutralize M KOH. (i) Choose an appropriate indicator for this titration. (ii) How many milliliters of KOH are needed for neutralization? A7: (i) Thymol blue or phenolphthalein (ii) a M a V a = b M b V b (1)(0.040)(15.0) = (1)(0.050)(X) X = 12.0 ml KOH

39 Q8: If represents HCl and if represent NaOH, then draw a particulate representation of the system halfway to neutralization if 1 M NaOH were being added to 1 M HCl. Label each particle in your drawing including any charge it may have. [Assume the diagram has 4 before the titration began.] A8: is H 2 O formed is Na 1+ is Cl 1- is H 1+ Q9: If the particulate representation below shows an acid-base system halfway to neutralization, then which of the following 1.0 M acids (HNO 3 or HC 2 H 3 O 2 ) was the one that was being neutralized by adding 1.0 M KOH? Explain. is H 2 O formed is K 1+ is OH 1- is H 1+ is unknown acid A9: HC 2 H 3 O 2 is the unknown acid. (1) must be a weak acid since it is in it molecular form. HC 2 H 3 O 2 is a weak acid and HNO 3 is a strong acid. (2) Halfway to neutralization, [ equals [ ]-acid anion which should be the case for a weak acid. ]-acid

40 Chapter-17 Lesson-9 VI Plotting an Acid-Base Titration Curve The profile of a titration curve can tell you a lot about the acid and base in the titration. Strong acid-strong base curves have no horizontal inflection points in either the lower or upper portion of the curve. Weak acid-strong base curves have a horizontal inflection point in the lower portion of the curve. Strong acid-weak base curves have a horizontal inflection point in the upper portion of the curve. The equivalence point occurs at the vertical inflection point on the curve. Polyprotic acids have two equivalence points on the curve ph 7 Equivalence Point ph 7 Equivalence Point 0 0 Base Added (ml) Base Added (ml) Adding a Strong Base to a Strong Acid Adding a Strong Base to a Weak Acid ph ph Equivalence Point 2 7 Equivalence Point 7 Equivalence Point Base Added (ml) Adding a Weak Base to a Strong Acid Base Added (ml) Adding Strong Base to a weak Polyprotic Acid

41 A) Plotting Titrations: Strong Acid with Strong Base Strong Acid (HX) + Strong Base (MOH) Neutral Salt + Water H + (aq) + X - (aq) + M + (aq) + OH - (aq) M + (aq) + X - (aq) + HOH Neutral salts do NOT react with water and thus do NOT change the ph of the water. 1 4 poh = -log [OH - ] ph = 14 poh ph 7 only neutral salt: ph = 7 excess strong base (OH - ) & neutral salt: excess mol OH - = mol OH - added initial mol H + 0 all strong acid (H + ): ph = -log [H + ] some strong acid (H + ) left & neutral salt: mol H + left = initial mol H + mol OH - added ph = -log [H + ] Strong Base Added It takes 5 points to plot this type of titration: (1) 0 ml of titrant added (2) 1 ml before neutralization (3) ml needed for neutralization (4) 1 ml after neutralization (5) double ml after neutralization Getting started: Calculate ml needed for neutralization: a M a V a = b M b V b Calculate initial moles of H 1+ to be titrated: a M a V a

42 Q1: Plot a titration curve for the neutralization of ml of a M HCl solution with M NaOH solution. A1: a M a V a = b M b V b (1)(0.0200)(15.00) = (1)(0.0150)(X) X = 20.0 ml NaOH needed for neutralization Initial mol H 1+ a M a V a = (1)(0.0200M)( L) = mol H 1+ (1) at 0.0 ml NaOH added (initial) ph = -log[hcl] = -log[0.0200] = (2) at 19.0 ml NaOH added (before neutralization) tot. vol. = 34.0 ml mol OH 1- added = (0.0190L)(0.0150M) = mol OH 1- mol H 1+ left = = mol H 1+ left [H 1+ left] = mol H L = M H 1+ left ph = -log[h 1+ left] = -log[ ] = (3) at 20.0 ml NaOH added (at neutralization) ph = (4) at 21.0 ml NaOH added (after neutralization) tot. vol. = 36.0 ml mol OH 1- excess = (0.0010L)(0.0150M) = mol OH 1- [OH 1- excess] = mol OH L = 4.17x10-4 M poh = -log[oh 1- excess] = -log[4.17x10-4 ] = 3.38 ph = 10.62

43 (5) at 40.0 ml NaOH added (after neutralization) tot. vol. = 55.0 ml mol OH 1- excess = (0.0200L)(0.0150M) = mol OH 1- [OH 1- excess] = mol OH L = M poh = -log[oh 1- excess] = -log[ ] = ph = ph ml of Base Added

44 Chapter-17 Lesson-10 B) Plotting Titrations: Weak Acid with a Strong Base Weak Acid (HA) + Strong Base (MOH) Basic Salt + Water HA (aq) + M + (aq) + OH - (aq) M + (aq) + A - (aq) + HOH Basic salts DO react with water and thus DO change the ph of the water. Basic salts RAISE the ph of the water. A - + HOH HA + OH - 14 only basic salt (A - ): mol A - = mol OH - added or excess strong base (OH - ) & basic salt: excess mol OH - = mol OH - added initial mol HA mol A - = initial mol HA poh = (pk b + log 1/[A - ]) poh = -log [OH - ] ph = 14 poh ph = 14 poh ph 7 ph = pk a + log [A - ]/[HA] where [A - ] = [HA] ph = pka some weak acid (HA) left & basic salt (A - ): mol HA left = initial mol HA mol OH - added mol A - = mol OH - added all weak acid (HA): ph = pk a + log [A - ]/[HA] 0 ph = (pk a + log 1/[HA]) Strong Base Added (ml)

45 It takes 7 points to plot this type of titration: (1) 0 ml of titrant added (2) 1 ml of titrant added (3) half ml needed for neutralization (4) 1 ml before neutralization (5) ml needed for neutralization (6) 1 ml after neutralization (7) double ml for neutralization Getting started: Calculate ml needed for neutralization: a M a V a = b M b V b Calculate initial moles of weak acid (HA) to be titrated. moles of H + initial = a M a V a Calculate pk a of weak acid. pk a = -logk a Calculate pk b of conjugate base of weak acid. pk b = 14 - pk a Q1: Plot a titration curve for the neutralization of ml of 2.00 M HF (K a = ) with 1.50 M NaOH. A1: a M a V a = b M b V b (1)(2.00)(15.00) = (1)(1.50)(X) X = 20.0 ml NaOH needed for neutralization Initial total mol H 1+ a M a V a = (1)(2.00M)( L) = mol H 1+ pk a = -log[k a ] = pk a = -log[ ] = 3.17 for HF pk b = 14 - pk a = = for F 1-

46 (1) at 0.0 ml NaOH added (initial) ph = ½[pK a + log(1/[hf])] = ½[ log(1/[2.00])] = 1.43 (2) at 1.0 ml NaOH added (before neutralization) mol OH 1- added = (0.0010L)(1.50M) = mol OH 1- added mol HF left = = mol HF left ph = pk a + log[f 1- ]/[HF] = log[0.0015*]/[ *] ph = 1.89 (3) at 10.0 ml NaOH added (½ before neutralization) ph = pk a = 3.17 (4) at 19.0 ml NaOH added (before neutralization) mol OH 1- added = (0.0190L)(1.50M) = mol OH 1- added mol HF left = = mol HF left ph = pk a + log[f 1- ]/[HF] = log[0.0285*]/[ *] ph = 4.45 (5) at 20.0 ml NaOH added (at neutralization) tot. vol. = 35.0 ml at neutralization all mol HF is now equal to mol F 1-. [F 1- ] = [0.0300mol L] = 0.857M poh = ½[pK b + log(1/[f 1- ])] = ½( log[1/0.857]) = 5.45 ph = 14 poh = = 8.55 (6) at 21.0 ml NaOH added (after neutralization) tot. vol. = 36.0 ml mol excess OH 1- = (0.0010L)(1.50M) = mol OH 1- excess [OH 1- excess] = mol L = M OH 1- excess poh = -log[oh 1- excess] = -log[0.042 M] = 1.38 ph = 12.62

47 (7) at 40.0 ml NaOH added (after neutralization) tot. vol. = 55.0 ml mol excess OH 1- = (0.0200L)(1.50M) = mol OH 1- excess [OH 1- excess] = mol L = M OH 1- excess poh = -log[oh 1- excess] = -log[0.545 M] = ph = ph ml of Base Added

48 Chapter-17 Lesson-11 C) Plotting Titrations: Weak Base with a Strong Acid Weak Base (B) + Strong Acid (HX) Acidic Salt B (aq) + H + (aq) + X - (aq) BH + (aq) + X - (aq) Acidic salts DO react with water and thus DO change the ph of the water. Acidic salts LOWER the ph of the water. BH + (aq) + HOH (l) B (aq) + H 3 O + (aq) 14 all weak base (B): poh = (pk b + log 1/[B]) ph = 14 poh some weak base (B) left & acidic salt (BH + ): mol B left = initial mol B mol H + added mol BH + = mol H + added poh = pk b + log [BH + ]/[B] ph = 14 poh ph 7 poh = pk b + log [BH + ]/[B] where [BH + ] = [B] poh = pkb (ph = 14 poh) 0 only acidic salt (BH + ): mol BH + = mol H + added or mol BH + = initial mol B ph = (pk a + log 1/[BH + ]) excess strong acid (H + ) & acidic salt: excess mol H + = mol H + added initial mol B ph = -log [H + ] Strong Acid Added (ml)