1 Applications of Aqueous Equilibria Chapter 15 AP Chemistry Seneca Valley SHS

2 The Common Ion Effect The solubility of a partially soluble salt is decreased when a common ion is added. Consider the equilibrium established when acetic acid, HC 2 H 3 O 2, is added to water. At equilibrium H + and C 2 H 3 O 2 - are constantly moving into and out of solution, but the concentrations of ions is constant and equal. If a common ion is added, e.g. C 2 H 3 O 2 - from NaC 2 H 3 O 2 (which is a strong electrolyte) then [C 2 H 3 O 2 - ] increases and the system is no longer at equilibrium. So, [H + ] must decrease.

3 The Common Ion Effect When a solution contains a salt having an ion common with one in equilibrium, the position of the equilibrium is driven away from the side containing that ion. Compare the pH values for sample problems 5 and 7 to see this!

4 Buffered Solutions Composition and Action of Buffered Solutions A buffer resists a change in pH when a small amount of OH - or H + is added. A buffer consists of a mixture of a weak acid (HX) and its conjugate base (X - ): The K a expression is

5 Buffered Solutions Composition and Action of Buffered Solutions When OH - is added to the buffer, the OH - reacts with HX to produce X - and water. But, the [HX]/[X - ] ratio remains more or less constant, so the pH is not significantly changed. When H + is added to the buffer, X - is consumed to produce HX. Once again, the [HX]/[X - ] ratio is more or less constant, so the pH does not change significantly.

6 Buffered Solutions Key Points on Buffered Solutions 1. They are weak acids or bases containing a common ion. 2.After addition of strong acid or base, deal with stoichiometry first, then equilibrium.

7 Buffered Solutions Addition of Strong Acids or Bases to Buffers We break the calculation into two parts: stoichiometric and equilibrium. The amount of strong acid or base added results in a neutralization reaction: X - + H 3 O +  HX + H 2 O HX + OH -  X - + H 2 O. By knowing how must H 3 O + or OH - was added (stoichiometry) we know how much HX or X - is formed.

8 Buffered Solutions Addition of Strong Acids or Bases to Buffers

9 Buffered Solutions Addition of Strong Acids or Bases to Buffers With the concentrations of HX and X - (note the change in volume of solution) we can calculate the pH from the Henderson-Hasselbalch equation Henderson-Hasselbalch Equation: For a particular buffering system, all solutions that have the same ratio [A - ]/[HA] will have the same pH.

10 Buffered Solutions Buffer Capacity and pH – Henderson- Hasselbach Equation If K a is small (i.e., if the equilibrium concentration of undissociated acid is close to the initial concentration), then

11 Buffered Solutions Buffer Capacity and pH Buffer capacity is the amount of acid or base neutralized by the buffer before there is a significant change in pH. Buffer capacity depends on the composition of the buffer. The greater the amounts of conjugate acid-base pair, the greater the buffer capacity. The pH of the buffer depends on K a.

12 Acid-Base Titrations Strong Acid-Base Titrations Strong Acid-Base Titrations Consider adding a strong base (e.g. NaOH) to a solution of a strong acid (e.g. HCl). –Before any base is added, the pH is controlled by the strong acid solution. Therefore, pH < 7. –When base is added, before the equivalence point, the pH is controlled by the amount of strong acid left in excess. Therefore, pH < 7. –At equivalence point, the amount of base added is stoichiometrically equivalent to the amount of acid originally present. Therefore, the pH is determined by the salt solution. Therefore, pH = 7.

13 Acid-Base Titrations Strong Acid-Base Titrations Strong Acid-Base Titrations Problems only involve a stoichiometry calculation. We know the pH at equivalent point is To detect the equivalent point, we use an indicator that changes color somewhere near The equivalence point in a titration is the point at which the acid and base are present in stoichiometric quantities. The end point in a titration is the observed point. The difference between equivalence point and end point is called the titration error.

14 Acid-Base Titrations Strong Acid-Base Titrations Strong Acid-Base Titrations

15 Acid-Base Titrations Weak Acid-Strong Base Titrations Weak Acid-Strong Base Titrations Consider the titration of acetic acid, HC 2 H 3 O 2 and NaOH. Before any base is added, the solution contains only weak acid. Therefore, pH is given by the equilibrium calculation. As strong base is added, the strong base consumes a stoichiometric quantity of weak acid: HC 2 H 3 O 2 (aq) + NaOH(aq)  C 2 H 3 O 2 - (aq) + H 2 O(l)

16 Acid-Base Titrations Weak Acid-Strong Base Titrations Weak Acid-Strong Base Titrations

17 Acid-Base Titrations Weak Acid-Strong Base Titrations Weak Acid-Strong Base Titrations There is an excess of acetic acid before the equivalence point. Therefore, we have a mixture of weak acid and its conjugate base. –The pH is given by the buffer calculation. First the amount of C 2 H 3 O 2 - generated is calculated, as well as the amount of HC 2 H 3 O 2 consumed. (Stoichiometry.) Then the pH is calculated using equilibrium conditions. (Henderson-Hasselbalch.)

18 Acid-Base Titrations Weak Acid-Strong Base Titrations Weak Acid-Strong Base Titrations At the equivalence point, all the acetic acid has been consumed and all the NaOH has been consumed. However, C 2 H 3 O 2 - has been generated. –Therefore, the pH is given by the C 2 H 3 O 2 - solution. –This means pH > 7. More importantly, pH  7 for a weak acid-strong base titration. After the equivalence point, the pH is given by the strong base in excess.

19 Acid-Base Titrations Weak Acid-Strong Base Titrations Weak Acid-Strong Base Titrations The inflection point is not as steep for a weak acid- strong base titration. The shape of the two curves after equivalence point is the same because pH is determined by the strong base in excess. Two features of titration curves are affected by the strength of the acid: –the amount of the initial rise in pH, and –the length of the inflection point at equivalence.

20 Acid-Base Titrations Weak Acid-Strong Base Titrations Weak Acid-Strong Base Titrations The weaker the acid, the smaller the equivalence point inflection. For very weak acids, it is impossible to detect the equivalence point.

21 Acid-Base Titrations Weak Acid-Strong Base Titrations Weak Acid-Strong Base Titrations Titration of weak bases with strong acids have similar features to weak acid-strong base titrations. Weak Acid-Strong Base Titrations: A Summary Step 1 - A stoichiometry problem - reaction is assumed to run to completion - then determine remaining species. Step 2 - An equilibrium problem - determine position of weak acid equilibrium and calculate pH.

22 Acid-Base Titrations Titrations of Polyprotic Acids Titrations of Polyprotic Acids In polyprotic acids, each ionizable proton dissociates in steps. Therefore, in a titration there are n equivalence points corresponding to each ionizable proton. In the titration of Na 2 CO 3 with HCl there are two equivalence points: –one for the formation of HCO 3 - –one for the formation of H 2 CO 3.

23 Indicator Color Change Another way determine the equivalence point of an acid-base titration is through the use of an acid-base indicator. Careful selection of the indicator will ensure that the end point is close to the equivalence point. Most common acid-base indicators are complex molecules that are themselves weak acids (HIn). They exhibit one color when the proton is attached and a different color when the proton is absent. Page 749 contains a list of acid-base indicators. Henderson-Hasselbach equation is very useful in determining the pH at which an indicator changes color.

24 Solubility Equilibria Solubility-Product Constant, K sp Solubility-Product Constant, K sp Consider for which K sp is the solubility product. (BaSO 4 is ignored because it is a pure solid so its concentration is constant.)

25 Solubility Equilibria Solubility-Product Constant, K sp Solubility-Product Constant, K sp In general: the solubility product is the molar concentration of ions raised to their stoichiometric powers. Solubility is the amount (grams) of substance that dissolves to form a saturated solution. Molar solubility is the number of moles of solute dissolving to form a liter of saturated solution.

26 Solubility Equilibria Solubility and K sp Solubility and K sp To convert solubility to K sp solubility needs to be converted into molar solubility (via molar mass); molar solubility is converted into the molar concentration of ions at equilibrium (equilibrium calculation), K sp is the product of equilibrium concentration of ions.

27 Solubility Equilibria Solubility and K sp Solubility and K sp

28 Solubility Equilibria Solubility and K sp Solubility and K sp Exercise Page 752 Copper (I) bromide has a measured solubility of 2.0x10 -4 mol/L at 25  C. Calculate its K sp value.

29 Solubility Equilibria Solubility and K sp Solubility and K sp Exercise Page 754 Calculate the K sp value for bismuth sulfide (Bi 2 S 3 ), which has a solubility of 1.0x mol/L at 25  C.

Solubility Equilibria Solubility and K sp Solubility and K sp Exercise Page 755 The Ksp value for copper (II) iodate is 1.4 x at 25º C. Calculate its solubility.

31 Solubility Equilibria Solubility and K sp Solubility and K sp A potassium chromate solution being added to aqueous silver nitrate, forming silver chromate.

32 Factors That Affect Solubility Common-Ion Effect Common-Ion Effect Solubility is decreased when a common ion is added. This is an application of Le Châtelier’s principle: as F - (from NaF, say) is added, the equilibrium shifts away from the increase. Therefore, CaF 2 (s) is formed and precipitation occurs. As NaF is added to the system, the solubility of CaF 2 decreases.

33 Factors That Affect Solubility Common-Ion Effect Common-Ion Effect

34 Factors That Affect Solubility Common-Ion Effect Common-Ion Effect Exercise Page 758 Calculate the solubility of solid CaF 2 (K sp = 4.0x ) in a M NaF solution.

35 Factors That Affect Solubility Solubility and pH Solubility and pH Again we apply Le Châtelier’s principle: –If the F - is removed, then the equilibrium shifts towards the decrease and CaF 2 dissolves. –F - can be removed by adding a strong acid: –As pH decreases, [H + ] increases and solubility increases. The effect of pH on solubility is dramatic.