Download presentation

Presentation is loading. Please wait.

Published byAnthony Shepherd Modified over 3 years ago

1
Chapter 15 Applying equilibrium

2
The Common Ion Effect l When the salt with the anion of a weak acid is added to that acid, l It reverses the dissociation of the acid. l Lowers the percent dissociation of the acid. l The same principle applies to salts with the cation of a weak base. l The calculations are the same as last chapter.

3
Buffered solutions l A solution that resists a change in pH. l Either a weak acid and its salt or a weak base and its salt. l We can make a buffer of any pH by varying the concentrations of these solutions. l Same calculations as before. l Calculate the pH of a solution that is.50 M HAc and.25 M NaAc (Ka = 1.8 x 10 -5 )

4
Na + is a spectator and the reaction we are worried about is HAc H + + Ac - l Choose x to be small x l We can fill in the table xx -x 0.50-x0.25+x Initial 0.50 M 0 0.25 M Final

5
l Do the math l Ka = 1.8 x 10 -5 1.8 x 10 -5 = x (0.25+x) (0.50-x) l Assume x is small = x (0.25) (0.50) x = 3.6 x 10 -5 l Assumption is valid l pH = -log (3.6 x 10 -5 ) = 4.44 HAc H + + Ac - x xx -x 0.50-x0.25+x Initial 0.50 M 0 0.25 M Final

6
Adding a strong acid or base l Do the stoichiometry first. –Use moles not molar l A strong base will grab protons from the weak acid reducing [HA] 0 l A strong acid will add its proton to the anion of the salt reducing [A - ] 0 l Then do the equilibrium problem. l What is the pH of 1.0 L of the previous solution when 0.010 mol of solid NaOH is added?

7
HAc H + + Ac - l In the initial mixture M x L = mol l 0.50 M HAc x 1.0 L = 0.50 mol HAc l Adding 0.010 mol OH - will reduce the HAc and increase the Ac - by 0.010 mole l Because it is in 1.0 L, we can convert it to molarity 0.50 mol 0.25 mol l 0.25 M Ac - x 1.0 L = 0.25 mol Ac - 0.49 mol 0.26 mol

8
HAc H + + Ac - l In the initial mixture M x L = mol l 0.50 M HAc x 1.0 L = 0.50 mol HAc l 0.25 M Ac - x 1.0 L = 0.25 mol Ac - l Adding 0.010 mol OH - will reduce the HAc and increase the Ac - by 0.010 mole l Because it is in 1.0 L, we can convert it to molarity 0.50 mol 0.25 mol 0.49 M 0.26 M

9
HAc H + + Ac - l Fill in the table 0.50 mol 0.25 mol 0.49 M 0.26 M HAc H + + Ac - x xx -x 0.49-x0.26+x Initial 0.49 M 0 0.26 M Final

10
l Do the math l Ka = 1.8 x 10 -5 1.8 x 10 -5 = x (0.26+x) (0.49-x) l Assume x is small = x (0.26) (0.49) x = 3.4 x 10 -5 l Assumption is valid l pH = -log (3.4 x 10 -5 ) = 4.47 HAc H + + Ac - x xx -x 0.49-x0.26+x Initial 0.49 M 0 0.26 M Final

11
Notice l If we had added 0.010 mol of NaOH to 1 L of water, the pH would have been. l 0.010 M OH - l pOH = 2 l pH = 12 l But with a mixture of an acid and its conjugate base the pH doesnt change much l Called a buffer.

12
General equation l Ka = [H + ] [A - ] [HA] l so [H + ] = Ka [HA] [A - ] l The [H + ] depends on the ratio [HA]/[A - ] l taking the negative log of both sides l pH = -log(Ka [HA]/[A - ]) l pH = -log(Ka)-log([HA]/[A - ]) l pH = pKa + log([A - ]/[HA])

13
This is called the Henderson- Hasselbach equation l pH = pKa + log([A - ]/[HA]) l pH = pKa + log(base/acid) l Works for an acid and its salt l Like HNO 2 and NaNO 2 l Or a base and its salt l Like NH 3 and NH 4 Cl l But remember to change K b to K a

14
l Calculate the pH of the following l 0.75 M lactic acid (HC 3 H 5 O 3 ) and 0.25 M sodium lactate (Ka = 1.4 x 10 -4 ) l pH = 3.38

15
l Calculate the pH of the following l 0.25 M NH 3 and 0.40 M NH 4 Cl (K b = 1.8 x 10 -5 ) l Ka = 1 x 10 -14 1.8 x 10 -5 l Ka = 5.6 x 10 -10 l remember its the ratio base over acid l pH = 9.05

16
Prove theyre buffers l What would the pH be if.020 mol of HCl is added to 1.0 L of both of the preceding solutions. l What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L of each of the proceeding. l Remember adding acids increases the acid side, l Adding base increases the base side.

17
Prove theyre buffers l What would the pH be if.020 mol of HCl is added to 1.0 L of preceding solutions. l 0.75 M lactic acid (HC 3 H 5 O 3 ) and 0.25 M sodium lactate (Ka = 1.4 x 10 -4 ) HC 3 H 5 O 3 H + + C 3 H 5 O 3 - Initially 0.75 mol 00.25 mol After acid 0.77 mol0.23 mol Compared to 3.38 before acid was added

18
Prove theyre buffers l What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L of the solutions. l 0.75 M lactic acid (HC 3 H 5 O 3 ) and 0.25 M sodium lactate (Ka = 1.4 x 10 -4 ) HC 3 H 5 O 3 H + + C 3 H 5 O 3 - Initially 0.75 mol 00.25 mol After acid 0.70 mol0.30 mol Compared to 3.38 before acid was added

19
Prove theyre buffers l What would the pH be if.020 mol of HCl is added to 1.0 L of preceding solutions. l 0.25 M NH 3 and 0.40 M NH 4 Cl K b = 1.8 x 10 -5 so Ka = 5.6 x 10 -10 NH 4 + H + + NH 3 Initially 0.40 mol 00.25 mol After acid 0.42 mol0.23 mol Compared to 9.05 before acid was added

20
Prove theyre buffers l What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L each solutions. l 0.25 M NH 3 and 0.40 M NH 4 Cl K b = 1.8 x 10 -5 so Ka = 5.6 x 10 -10 NH 4 + H + + NH 3 Initially 0.40 mol 00.25 mol After acid 0.35 mol0.30 mol Compared to 9.05 before acid was added

21
Buffer capacity l The pH of a buffered solution is determined by the ratio [A - ]/[HA]. l As long as this doesnt change much the pH wont change much. l The more concentrated these two are the more H + and OH - the solution will be able to absorb. l Larger concentrations = bigger buffer capacity.

22
Buffer Capacity l Calculate the change in pH that occurs when 0.020 mol of HCl(g) is added to 1.0 L of each of the following: l 5.00 M HAc and 5.00 M NaAc l 0.050 M HAc and 0.050 M NaAc l Ka= 1.8x10 -5 l pH = pKa

23
Buffer Capacity l Calculate the change in pH that occurs when 0.040 mol of HCl(g) is added to 1.0 L of 5.00 M HAc and 5.00 M NaAc l Ka= 1.8x10 -5 HAc H + + Ac - Initially 5.00 mol 05.00 mol After acid 5.04 mol4.96 mol Compared to 4.74 before acid was added

24
Buffer Capacity l Calculate the change in pH that occurs when 0.040 mol of HCl(g) is added to 1.0 L of 0.050 M HAc and 0.050 M NaAc l Ka= 1.8x10 -5 HAc H + + Ac - Initially 0.050 mol 0 0.050 mol After acid 0.090 mol 0 0.010 mol Compared to 4.74 before acid was added

25
Buffer capacity l The best buffers have a ratio [A - ]/[HA] = 1 l This is most resistant to change l True when [A - ] = [HA] l Makes pH = pKa (since log 1 = 0)

26
Titrations l Millimole (mmol) = 1/1000 mol l Molarity = mmol/mL = mol/L l Makes calculations easier because we will rarely add liters of solution. l Adding a solution of known concentration until the substance being tested is consumed. l This is called the equivalence point. l Graph of pH vs. mL is a titration curve.

27
Titration Curves

28
pH mL of Base added 7 l Strong acid with strong Base l Equivalence at pH 7

29
pH mL of Base added >7 l Weak acid with strong Base l Equivalence at pH >7 l When the acid is neutralized it makes a weak base 7

30
pH mL of acid added 7 l Strong base with strong acid l Equivalence at pH 7

31
pH mL of acid added <7 l Weak base with strong acid l Equivalence at pH <7 l When the base is neutralized it makes a weak acid 7

32
Strong acid with Strong Base l Do the stoichiometry. l mL x M = mmol l There is no equilibrium. l They both dissociate completely. l The reaction is H + + OH - HOH l Use [H + ] or [OH - ] to figure pH or pOH l The titration of 50.0 mL of 0.200 M HNO 3 with 0.100 M NaOH

33
Weak acid with Strong base l There is an equilibrium. l Do stoichiometry. –Use moles l Determine major species l Then do equilibrium. l Titrate 50.0 mL of 0.10 M HF (Ka = 7.2 x 10 -4 ) with 0.10 M NaOH

34
Summary l Strong acid and base just stoichiometry. l Weak acid with 0 ml of base - K a l Weak acid before equivalence point –Stoichiometry first –Then Henderson-Hasselbach l Weak acid at equivalence point- K b -Calculate concentration l Weak acid after equivalence - leftover strong base. -Calculate concentration

35
Summary l Weak base before equivalence point. –Stoichiometry first –Then Henderson-Hasselbach l Weak base at equivalence point K a. -Calculate concentration l Weak base after equivalence – left over strong acid. -Calculate concentration

36
Indicators l Weak acids that change color when they become bases. l weak acid written HIn l Weak base l HIn H + + In - clear red l Equilibrium is controlled by pH l End point - when the indicator changes color. l Try to match the equivalence point

37
Indicators l Since it is an equilibrium the color change is gradual. l It is noticeable when the ratio of [In - ]/[HI] or [HI]/[In - ] is 1/10 l Since the Indicator is a weak acid, it has a Ka. l pH the indicator changes at is. l pH=pKa +log([In - ]/[HI]) = pKa +log(1/10) l pH=pKa - 1 on the way up

38
Indicators l pH=pKa + log([HI]/[In - ]) = pKa + log(10) l pH=pKa+1 on the way down l Choose the indicator with a pKa 1 more than the pH at equivalence point if you are titrating with base. l Choose the indicator with a pKa 1 less than the pH at equivalence point if you are titrating with acid.

39
Solubility Equilibria Will it all dissolve, and if not, how much?

40
l All dissolving is an equilibrium. l If there is not much solid it will all dissolve. l As more solid is added the solution will become saturated. l Soliddissolved l The solid will precipitate as fast as it dissolves. l Equilibrium

41
General equation l M + stands for the cation (usually metal). l Nm - stands for the anion (a nonmetal). l M a Nm b (s) aM + (aq) + bNm - (aq) l K = [M + ] a [Nm - ] b /[M a Nm b ] l But the concentration of a solid doesnt change. l K sp = [M + ] a [Nm - ] b l Called the solubility product for each compound.

42
Watch out l Solubility is not the same as solubility product. l Solubility product is an equilibrium constant. l it doesnt change except with temperature. l Solubility is an equilibrium position for how much can dissolve. l A common ion can change this.

43
Calculating K sp l The solubility of iron(II) oxalate FeC 2 O 4 is 65.9 mg/L l The solubility of Li 2 CO 3 is 5.48 g/L

44
Calculating Solubility l The solubility is determined by equilibrium. l Its an equilibrium problem. l Watch the coefficients l Calculate the solubility of SrSO 4, with a Ksp of 3.2 x 10 -7 in M and g/L. l Calculate the solubility of Ag 2 CrO 4, with a Ksp of 9.0 x 10 -12 in M and g/L.

45
Relative solubilities l Ksp will only allow us to compare the solubility of solids that fall apart into the same number of ions. l The bigger the Ksp of those the more soluble. l If they fall apart into different number of pieces you have to do the math.

46
Common Ion Effect l If we try to dissolve the solid in a solution with either the cation or anion already present less will dissolve. l Calculate the solubility of SrSO 4, with a Ksp of 3.2 x 10 -7 in M and g/L in a solution of 0.010 M Na 2 SO 4. l Calculate the solubility of SrSO 4, with a Ksp of 3.2 x 10 -7 in M and g/L in a solution of 0.010 M SrNO 3.

47
pH and solubility l OH - can be a common ion. l More soluble in acid. l For other anions if they come from a weak acid they are more soluble in a acidic solution than in water. l CaC 2 O 4 Ca +2 + C 2 O 4 -2 l H + + C 2 O 4 -2 HC 2 O 4 - l Reduces [C 2 O 4 -2 ] in acidic solution.

48
Precipitation l Ion Product, Q =[M + ] a [Nm - ] b l If Q>Ksp a precipitate forms. l If Q

49
Selective Precipitations l Used to separate mixtures of metal ions in solutions. l Add anions that will only precipitate certain metals at a time. l Used to purify mixtures. l Often use H 2 S because in acidic solution Hg +2, Cd +2, Bi +3, Cu +2, Sn +4 will precipitate.

50
Selective Precipitation l Then add OH - solution [S -2 ] will increase so more soluble sulfides will precipitate. l Co +2, Zn +2, Mn +2, Ni +2, Fe +2, Cr(OH) 3, Al(OH) 3

51
Selective precipitation l Follow the steps l First with insoluble chlorides (Ag, Pb, Ba) l Then sulfides in Acid. l Then sulfides in base. l Then insoluble carbonate (Ca, Ba, Mg) l Alkali metals and NH 4 + remain in solution.

52
Complex ion Equilibria l A charged ion surrounded by ligands. l Ligands are Lewis bases using their lone pair to stabilize the charged metal ions. l Common ligands are NH 3, H 2 O, Cl -,CN - l Coordination number is the number of attached ligands. l Cu(NH 3 ) 4 2+ has a coordination # of 4

53
The addition of each ligand has its own equilibrium l Usually the ligand is in large excess. l And the individual Ks will be large so we can treat them as if they go to completion. l The complex ion will be the biggest ion in solution.

54
l Calculate the concentrations of Ag +, Ag(S 2 O 3 ) -, and Ag(S 2 O 3 ) -3 in a solution made by mixing 150.0 mL of 0.010 M AgNO 3 with 200.0 mL of 5.00 M Na 2 S 2 O 3 l Ag + + S 2 O 3 -2 Ag(S 2 O 3 ) - K 1 =7.4 x 10 8 l Ag(S 2 O 3 ) - + S 2 O 3 -2 Ag(S 2 O 3 ) 2 -3 K 2 =3.9 x 10 4

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google