The Common Ion Effect and Acid/Base Equilibria

Transcription

1 The Common Ion Effect and Acid/Base Equilibria

2 Additional Aspects of Acid/Base Equilibria So far, we have examined the equilibrium concentrations of ions in solutions containing a weak acid or weak base Now, we will consider solutions that contain a weak acid and a soluble salt of that acid

3 Additional Aspects of Acid/Base Equilibria An example of a solution containing a weak acid and soluble salt is acetic acid and sodium acetate Sodium acetate is a soluble ionic compound and therefore, a strong electrolyte (dissociates completely): CH 3 COONa aq Na + aq + CH 3 COO aq Acetic acid is a weak electrolyte that ionizes only partially, and is represented by a the dynamic equilibrium: CH 3 COOH aq H + aq + CH 3 COO aq

4 So, What is the Common-Ion Effect? If we add sodium acetate to a solution of acetic acid in water, the CH 3 COO - from CH 3 COONa causes the equilibrium concentrations of the substances in the acetic acid system to shift to the left, thereby decreasing the equilibrium concentration of H + CH 3 COONa (s) CH 3 COOH (aq) Na + (aq) + CH 3 COO - (aq) H + (aq) + CH 3 COO - (aq) Common Ion In other words, whenever a weak electrolyte and a strong electrolyte containing a common ion are together in solution, the weak electrolyte ionizes less than it would if it were alone in solution We call this observation the common-ion effect

5 A General Overview of the Common-Ion Effect The common-ion effect is generally defined as the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance The presence of a common ion suppresses the ionization of a weak acid or a weak base

6 Real-Life Applications of the Common-Ion Effect The most important application of acid-base solutions containing a common ion is for buffering A buffered solution is one that resists a change in its ph when either hydroxide ions or protons are added This is because it contains both an acid to neutralize added OH - ions and a base to neutralize added H + ions HUGE EXAMPLE OF A BUFFERED SOLUTION Our blood! Blood can absorb the acids and bases produced in biologic reactions without changing its ph Vital because cells can survive only in a very narrow ph range! Buffering system in blood involves HCO 3 - and H 2 CO 3

7 Composition of a Buffer Solution A buffer solution can consist of: A weak acid and its conjugate base (its salt) HF and NaF A weak base and its conjugate acid (its salt) NH 3 and NH 4 Cl

8 Ways to Make a Buffered Solution There are two ways to make a buffer: Mix a weak acid or weak base with a salt of that acid or base For example, the CH 3 COOH/CH 3 COO buffer can be prepared by adding CH 3 COONa to a solution of CH 3 COOH Make the conjugate acid or base from a solution of a weak base or acid by the addition of a strong acid or base to neutralize about half of the weak acid or weak base For example, the CH 3 COOH/CH 3 COO buffer can be prepared by adding some NaOH to the solution enough to neutralize about half of CH 3 COOH according to the reaction: CH 3 COOH aq + OH aq CH 3 COO aq + H 2 O (l)

9 Practice! # 4 A N D # 6 O N P A G E

10 ph and Buffer Solutions Because a buffer resists changes in ph, it is useful to calculate the initial ph of the solution So, how do we do this?

11 Steps for Calculating ph of Buffer Solutions 1. THINK Which solutes are strong electrolytes and which are weak electrolytes? 2. THINK Major species in solution 3. THINK What is the important equilibrium reaction that is the source of H + (therefore, controls ph)? 4. WRITE Equilibrium equation 5. THINK Do not know equilibrium conditions so use ICE table and K a expression to solve for [H + ] ([H 3 O + ] ) Note there is an initial concentration of conjugate base in the ICE table! Make assumption that [HA] int x = [HA] eq = [HA] int if K a << 1 1. WRITE ph = -log[h + ]

12 Practice! # 8 O N P A G E

13 There s an Even Easier Way to Calculate ph of a Buffer! The acid dissociation equilibrium expression can be rearranged to provide a useful equation when calculating ph: K a [ H ][ A ] [ H K HA ] [ ] a [ HA] [ A ] HA log[ H ] log Ka log [ ] [ A ]

14 Derivation Continued ph = pk a + log A HA = pk a + log base acid This log form of the expression for K a is called the Henderson-Hasselbach equation It is useful for calculating the ph of solutions when the ratio [HA]/[A-] is known AND K a <<< 1

15 Practice! # 9 O N P A G E

16 Preparing a Buffer In the laboratory, it is useful to know the amounts of the acid and its conjugate base or weak base and conjugate acid needed to achieve a specific ph The most effective buffer occurs when the ratio of [A - ] to [HA] or [B] to [BH + ] is 1 If this ratio is 1, then according to the Henderson-Hasselbalch equation: ph = pk a

17 Preparing a Buffer with a Certain ph Method #1 To do this: 1. THINK Major species in buffer solution 2. THINK What conjugate acid-base pair determines ph? 3. WRITE Equilibrium equation between conjugate acid-base pair and water 4. WRITE Equilibrium expression 5. WRITE Obtain [H + ] or [OH - ] using given ph: [H + ] = 10 -ph or [OH - ] = 10 -poh 1. WRITE Plug and chug using all known information into equilibrium expression and solve for unknown! 1. Usually, this includes [HA] or [B] and K a /K b value

18 Preparing a Buffer Solution with a Specific ph Method #2 Choose a weak acid whose pk a is close to the desired ph You can: Calculate [H + ] from ph Substitute this value with K a value of weak acid into the equation: H + = K a [HA] [A ] Choose the ratio [HA]/[A-] that is closest to 1 OR: Substitute pk a (-log K a ) and ph values into the Henderson- Hasselbach Equation This will give a ratio of [A-]/[HA] Then, convert ratio to molar quantities

19 Practice! # 1 4 O N P A G E

20 So, Why are Buffers So Important? Buffers find many important applications in the laboratory and in medicine Many biological reactions occur at the optimal rates only when properly buffered Remember, buffered solutions resist ph change with addition of a STRONG acid or base The amount of acid or base that buffer can neutralize before the ph begins to change significantly is called the buffer capacity

21 What Exactly Happens When a Strong Acid or Base is Added to a Buffer? For a weak acid buffer (ph < 7) Added strong acid (H + ) reacts with conjugate base Added strong base (OH - ) reacts with weak acid For a weak base buffer (ph > 7): Added strong acid (H + ) reacts with weak base (B) Added strong base (OH - ) reacts with conjugate acid (BH + )

22 Buffer Action Acetic Acid/Acetate Buffer

23 Buffer Action Buffer Response to Strong Acid Buffer Response to Strong Base

24 Calculating How ph of a Buffer Responds to Addition of Strong Acids and Bases Solution #1 Do a Before Reaction Change After Reaction table using MOLES ONLY for stoichiometry calculations to determine new moles of buffer components Assume reaction with H + /OH - goes to completion Volume of solution changed with addition of SA/SB so calculate new concentrations by assuming volumes are additive Set-up an ICE table with new concentrations to determine equilibrium concentrations Calculate ph from [H+] using K a expression Solution #2 Do a Before Reaction Change After Reaction table for stoichiometry calculations to determine new moles of buffer components Assume reaction with H + /OH - goes to completion Volume of solution changed with addition of SA/SB so calculate new concentrations by assuming volumes are additive Use Henderson-Hasselbach equation to solve for ph

25 Steps to Calculate the ph of a Buffer After Addition of Strong Acid or Base

26 Buffer Tutorial How to Do Buffer Calculations

27 Practice! # 1 3 O N P A G E

28 Making a Buffer Calculations You want to prepare ml of a buffer with a ph = 10.00, with both the acid and conjugate base having molarities between 0.10 M to 1.00 M. You may choose from any of the acids listed below: Name Formula Ka1 Ka2 Ka3 acetic acid HC2H3O2 1.8 x 10-5 tartaric acid H2C4H4O6 1.0 x x 10-5 citric acid H3C6H5O7 7.4 x x x 10-7 malonic acid H2C3H2O4 1.5 x x 10-6 carbonic acid H2CO3 4.3 x x phosphoric acid H3PO4 7.5 x x x ammonium NH x You must select an acid with a K a value close to 10 - assigned ph. The only two options are ammonium or the hydrogen carbonate ions. K a = 5.6 x = H+ NH 3 + NH 4 K a2 = 5.6 x = H+ CO HCO 3

29 K a = 5.6 x = H+ NH x = NH 3 + NH 4 + NH 4 Making a Buffer Calculations 5.6 x H + = NH = NH 3 + NH 4 Divide both [ ] by = NH 3 + NH 4 + NH mol NH x g NH 4 Cl = 500 ml g NH 4Cl + = 2.68 g NH 4 Cl 1000 ml 1 mol NH mol NH x ml conc NH 3 = 500 ml buffer ml conc NH ml buff 14.8 mol NH 3 =18.9 ml 14.8 M NH 3

30 Making a Buffer Calculations Place ~250 ml of distilled water in a 500 ml volumetric flask. Add 2.68 g NH 4 Cl and dissolve. Then mix in 18.9 ml of 14.8 M NH 3. Fill with distilled water to the 500 ml mark on the flask. If there is no concentrated NH 3 available, the NH 3 can be produced by neutralizing additional NH 4 Cl with 1.00 M NaOH mol NH x g NH 4 Cl = 500 ml buffer 3 1 mol NH 4Cl 58.5 g NH 4Cl 1000 ml buff 1 mol NH 3 1 mol NH 4 Cl =16.38 g NH 4 Cl x ml NaOH = 500 ml buffer = 280. ml NaOH 0.56 mol NH 3 1 mol NaOH 1000 ml NaOH 1000 ml buff 1 mol NH mol NaOH Dissolve g NH 4 Cl (2.68 g g) in 280. ml of 1.00 M NaOH. Then dilute with distilled water and fill to the 500 ml mark on the flask.

31 Making a Buffer Calculations Place ~250 ml of distilled water in a 500 ml volumetric flask. Add 2.68 g NH 4 Cl and dissolve. Then mix in 18.9 ml of 14.8 M NH 3. Fill with distilled water to the 500 ml mark on the flask. If there is no NH 4 Cl available, the NH + 4 can be produced by neutralizing additional NH 3 with 1.00 M HCl mol NH x ml NH 3 = 500 ml buffer 4 1 mol NH ml NH ml buff 1 mol NH mol NH 3 = 3.38 ml NH mol NH x ml HCl = 500 ml buffer 4 1 mol HCl 1000 ml HCl 1000 ml buff 1 mol NH mol HCl = 50.0 ml HCl Place ~250 ml distilled water in volumetric flask. Add 50.0 ml of 1.00 M HCl and mix. Then add 22.3 ml of concentrated NH 3 (18.9 ml ml). Mix and fill with distilled water to the 500 ml mark on the flask.

32 Making a Buffer Calculations K a2 = 5.6 x = H+ CO HCO x H + = CO HCO = CO x HCO 3 1 = CO HCO Multiply both [ ] by = CO HCO 3 Prepare 500. ml of the buffer that has [CO 2-3 ] = M and [HCO 1-3 ] = M.

33 Making a Buffer Calculations Prepare 500. ml of the buffer that has [CO 3 2- ] = M and [HCO 3 1- ] = M. x g Na 2 CO 3 = 500 ml mol CO g Na 2CO ml 1 mol CO 3 = 5.30 g Na 2 CO 3 x g NaHCO 3 = 500 ml mol HCO g NaHCO ml 1 mol HCO 3 = 7.52 g NaHCO 3 Place ~250 ml distilled water in a 500 ml volumetric flask. Add 5.30 g Na 2 CO 3 and 7.52 g NaHCO 3 and dissolve. Fill with distilled water to the 500 ml mark on the flask.