Acid-Base Solutions - Applications 1 The Common Ion Effect Consider the equilibrium established when acetic acid, HC 2 H 3 O 2, is added to water. CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) What does Le Chatelier s Principle tell me if I add CH 3 COO - to the solution How does the [H 3 O + ] change? How does the ph change? 2 Mixing Salts and Weak Acids or Bases The Common Ion Effect What happens when one adds two salts together in a solution if both salts contain the same anion? Ex. Mix NaCH 3 COO and KCH 3 COO CH 3 COO - (aq) + H 2 O (aq) CH 3 COOH(aq) + OH - (aq) So what happens when one places mixes a salt and an acid or a base in solution? Ex Mix CH 3 COOH and NaCH 3 COO CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) This exhibits the common ion effect. 3 1
Calculate the ph of a 0.100 M soln of acetic acid, CH 3 COOH. You should be able to do this, the ph is 2.879 Suppose one mixes a 0.100 M solution of acetic acid with a 0.0500M solution of sodium acetate. What is the ph of the solution? 4 Suppose I mix a.100 M solution of acetic acid with a 0.0500M solution of sodium acetate. What is the ph of the solution? K a = 1.74 x 10-5 CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) I 0.100 0.0500 C -x +x +x E 0.100 -x 0.0500 +x +x 5 Suppose I mix a.100 M solution of acetic acid with a 0.0500M solution of sodium acetate. What is the ph of the solution? K a = 1.74 x 10-5 (0.0500 x)(x) 5 1.74 x 10 (0.100 x) x [H O ] 3.48 x 10 3 ph 4.458 without the common ion, ph 2.789-5 6 2
An Application of the Common Ion Effect Buffers Solutions A buffer is defined as a solution with a common ion that resists a change in ph when a small amount of OH - or H + is added. One example of a buffer is a mixture of a weak acid (HX) and its conjugate base (X - ). There is a common ion in this case the conjugate base. 7 How do Buffers Work? I have a source of a weak acid in the solution: CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) What happens if I add base? The /[X - ] ratio remains more or less constant, so the ph is not significantly changed. 8 The ph is Controlled by the Acid/Conjugate base Ratio - [H ][X ] Ka [H ] K a - [X ] Now take the log of both sides 9 3
K a - [H ][X ] Now take the log of both sides and rearrange the log[h ] log K expression. a - [X ] log[h ]. K a - [X ] log 10 Henderson-Hasselbalch (H-H) equation for a buffer that consists of a weak acid and its conjugate base : ph pk a - [A ] log [HA] 11 Henderson-Hasselbalch (H-H) equation for Buffers If one knows the HX and X - concentrations after the addition of strong acid or base, we can calculate the ph of the resulting solution from the Henderson-Hasselbalch equation for the equilibrium component of the problem. Since the volume doesn t change, the volume cancels out. ph p K ph p K a a - [X ] log moles of conjugate base log moles of conjugate acid 12 4
For a basic buffer containing a weak base and it s conjugate acid: poh pk b [BH ] log [B] 13 Assumptions made to maintain a relatively constant ph by a buffer: The number of moles of weak acid and its conjugate base must be large in comparison to any amount of additional acid or base supplied to the solution. The ratio of the weak acid to conjugate base should be roughly 1:1. Buffer capacity will be okay if [HA] is at least 20 x K a. Five percent rule must apply or the H-H equation won t work. Then you need to use an ICE table. 14 Buffer Capacity 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 acidbase pair, the greater the buffer capacity. However, the ph of the buffer still depends on K a. 15 5
Buffer Capacity The acid buffer range is 0.1 < [A - ] <10 [HA] and similarly l for the base. 16 What happens when some strong acid or base is added to a buffer? Think of buffer problems in two parts when an acid or base is added: 1.The stoichiometry of the reaction if a strong acid or base is added to the buffer 2.The equilibrium calculation for the reaction 17 How Buffers Work 18 6
First, the stoichiometry. If you have an acidic buffer, (weak acid and it s conjugate base) you will need to calculate the # millimoles (mmol) of each species in solution. Will also need to know the # of mmol of acid or base added. Recall M x V(mL) = # mmol Need to calculate l the millimoles l of weak acid and millimoles of conjugate base in solution. 19 A buffer contains 25.00 ml of.100 M solution of acetic acid and 25.00 ml of 0.0500M solution of sodium acetate. What is the ph? pk a = 4.74 Need to solve the stoichiometry first. For a weak acid/conjugate base buffer problem, write down the weak acid ionization reaction to keep track of the stoichiometry. t 20 A buffer contains 25.00 ml of 0.100 M solution of acetic acid and 25.00 ml of 0.0500M solution of sodium acetate, what is the ph of the solution? pka = 4.74 Stoichiometry first: How many mmols of acetic acid? 2.5 mmol How many mmoles of acetate ion? 1.25 mmol Now use the H-H eqn for the equilibrium part: Now use H-H Eqn: pka = 4.74 ph 4.74 1.25 log 4.439 2.50 21 7
If 1.00 ml of 0.050 M HCl solution is added to a buffer containing 25.00 ml of 0.100 M solution of acetic acid and 25.00 ml of 0.0500M solution of sodium acetate, what is the ph of the solution? pka = 4.74 Stoichiometry first: 1. How many mmols of acetic acid before reaction? 2. How many mmoles of acetate ion before reaction? 3. What does the HCl react with? 4. How many mmols of HCl react? 5. How many mmols of acetic acid after reaction? 6. How many mmoles of acetate ion after reaction? 22 If 1.00 ml of 0.0500 M HCl solution is added to a buffer containing 25.00 ml of 0.100 M solution of acetic acid and 25.00 ml of 0.0500M solution of sodium acetate, what is the ph of the solution? pka = 4.74 How many mmols of acetic acid before reaction? 2.5 mmol How many mmoles of acetate ion before reaction? 1.25 mmol What does the HCl react with? Acetate ion How many mmols of HCl react? 0.0500 mmol How many mmols of acetic acid after reaction?2.5+0.05 = 2.55 mmol How many mmoles of acetate ion after reaction? 1.25-0.05=1.20 mmol Now use H-H Eqn: pka = 4.74 ph = 4.41 1.20 mmol ph 4.74 log 4.413 2.55 mmol 23 Suppose I add 1.00 ml of 0.0500 M HCl to 50.00-mL of DI water. What is the ph of the solution? 1.00 ml x 0.0500 M 9.809 x 10 51.00 100mL -4 3 [H O ] ph 3.009 24 8
Compare using 3 sig fig s ph of buffer in 50 ml = 4.44 ph of fbuffer with 1 ml of HCl added d= 4.41 ph of 1 ml of HCl in 50 ml DI = 3.01 You can see that the buffer resists the change in the ph. 25 Example of Basic Buffer Calculate the ph of a solution created by mixing 200.0 ml of 0.400 M ammonia and 200.0 ml of 1.00 M ammonium chloride. Hint: The instant the two solutions are mixed what are the initial concentrations of each? Calculate the ph of Calculate the ph of a solution created by mixing 200.0 ml of 0.400 M ammonia and 200.0 ml of 1.00 M ammonium chloride when 15.00 ml of 0.100 M HI is added to the solution. 26 9