Buffer Solutions Buffers are solutions with the ability to resist the addition of strong acids or strong bases, within limits.
They play an important role in chemical processes where it is essential that a fairly constant pH is maintained. For example, the pH of your blood lies at about 7.35. If this value drops below 7.0 (acidosis) the results are fatal. Also if it rises above 7.7 (alkalosis) the results are also fatal. Fortunately our blood contains a buffering system which maintains the acidity at the proper level. If it were not for the protection of the buffering system, we could not eat and adsorb many of the acidic fruit juices and foods in our diet.

Buffers and the Henderson-Hasselbalch Equation
-many biological processes generate or use H + - the pH of the medium would change dramatically if it were not controlled (leading to unwanted effects) --biological reactions occur in a buffered medium where pH changes slightly upon addition of acid or base -most biologically relevant experiments are run in buffers
how do buffered solutions maintain pH under varying conditions?
to calculate the pH of a solution when acid/base ratio of weak acid is varied: Henderson-Hasselbalch equation comes from: K a = [H + ] [A – ] / [HA] take ( – log) of each side and rearrange, yields:
pH = pK a + log ( [A – ] / [HA] )
some examples using HH equation: what is the pH of a buffer that contains the following?
1 M acetic acid and 0.5 M sodium acetate

Blood buffer is made up from the dissolved carbon dioxide in the plasma. CO 2 (g) + H 2 O <-------> H 2 CO 3 <------> HCO 3 + H 3 O + When a base is added it reacts with the carbonic acid. OH + H 2 CO 3 <--------> HCO 3 + H 2 O When an acid is added it reacts with the bicarbonate ion. H 3 O + + HCO 3 <---------> H 2 CO 3 + H 2 O

There are 3 basic types of calculations that can be done with buffer systems and these will be covered next.
1.
The pH of a Buffered Solution Because HA is a weak acid, very little of it is dissociated at equilibrium. Even if HA were the only solute. But the solution also contains A from the dissolved salt. The presence of A suppresses the already slight ionization of HA by shifting the following equilibrium to the left. You should recognize this for the common ion effect , which it is.

HA <--------> H + + A ie. The value of [HA] at equilibrium = [HA] initially and the value of [A ] at equilibrium = [A ] initially. Therefore we can make 2 assumptions: [HA] equilibrium = [HA] from the initial [acid] = [acid] [A ] equilibrium = [A ] from the initial [salt] = [anion] Therefore we can do this: [H + ] = K a x [acid] [anion] If we take the -log of both sides: -log[H + ] = -log K a + (-log [ acid] ) [anion]

Two factors govern the pH in a buffered solution. (1) pK a of the weak acid; (2) ratio of the initial molar [ ]'s of the acid and it's salt. If a solution has [anion] = [acid] then the log [anion] = log 1 = 0 [acid] Therefore the pH of the solution will turn out equal to the pKa of the weak acid. What mostly determines where on a pH scale a buffer can work best is the pK a of the weak acid. Then by adjusting the ratio of [anion] to [acid], we can cause shifts so that the pH of the buffered solution comes out on one side or the other of this value of pH.

Easy Problem : Calculating the pH of a Buffered Solution To study the effect of a weakly acidic culture medium on the growth of a certain strain of bacterium, a microbiologist prepared a buffer solution by making it with 0.11 mol/L NaCH 3 COO, sodium acetate, and also 0.09 mol/L CH 3 COOH, acetic acid, What is the final pH? K a for acetic acid = 1.8 x 10 -5 Therefore pK a = 4.74 [acid] = 0.090 mol/L [anion] = 0.11 mol/L pH = 4.74 + log (0.11) = 4.74 + log 1.2 = 4.74 + 0.079 = 4.82
(0.9)

The Effectiveness of a Buffer (Important calculation) Suppose we drop 0.01 moles of strong base into our buffer from the last example. What will be the measured effects? HA start 0.09 M + OH ---- <-------> A + 0.11 M finish -0.01 M 0.08 M pH = pK a -0.1 M 0.0 M + log [anion] [acid] H 2 O +0.01 M ( These are changes) 0.12 M = 4.74 + log (0.12) (0.08) = 4.74 + 0.18 = 4.92
The change in pH is small compared to what it would have been in pure water! (Do the calculation !!!)

Buffer Capacity No buffer has an unlimited capacity. ie; buffers can only absorb so much abuse before they are destroyed. The capacity of a buffer is the amount of acid or base it can handle before the pH of the solution changes drastically. If you add enough strong acid to neutralize all of the buffers basic component, then additional strong acid will make the pH drop rapidly. The same applies for a strong base and the buffers acidic component. The buffer's pH is a function of it's pK a and the ratio of concentrations of anion and acid, but the buffer's capacity depends upon actual concentrations.

Preparation of A Buffer To prepare a buffer first choose an acid with a pK a within a ±1 pH unit of the desired value. Then manipulate the ratios to get the desired pH. A solution buffered at pH 5.00 is needed in a chemistry experiment. Can we use acetic acid and sodium acetate to make it? If so, what ratio of acetate ion to acetic acid is needed? K a of acetic acid = 1.8 x 10 -5 the pK a pK a = 4.74 is within a range of 5 ± 1 i.e. between the values of 4 - 6. therefore, acetic acid is okay to use. next use pH = pK a + log [anion] [acid] so 5.00 = 4.74 + log [anion] acid] 10 0.26 = [anion] = 1.8
[acid] Therefore the ratio of anion to acid is 1.8 to 1. Therefore, use a ratio of 1.8 moles of acetate ion to 1.0 mole of acetic acid. The solution will then be buffered at a pH of 5.00 OR We could use a ratio of 0.18 moles to 0.100 moles if we wanted a smaller buffering capacity.

A Basic Buffer: NH 3 + H 2 O <-------> NH 4 + + OH Taking the -log of both sides A generalized version of this basic Henderson Hasselbach equation is thus

Base buffer problem: A chemistry student needs 250 mL of a solution buffered at a pH of 9.00. How many grams of ammonium chloride have to be added to 250 mL of 0.2 mol/L NH 3 assumed not to change.) pH = 9.0
up ) to make such a buffer? (Volume is pOH = 5.0 [base] = 0.2
pK b of ammonia = 4.74 (look pOH = pK b +log [cation] [base] 5.00 = 4.74 + log [cation] (0.2) [cation] = 10 0.26
X 0.2 = 1.8 * 0.2 = 0.36 mol/L of the NH 4 + . But we only need enough for 250 mL so 0.36 mol = __x___ 1000 mL 250 mL so x = 0.09 moles of NH 4 + ions needed Since the NH 4 + comes from NH 4 Cl then we also need 0.09 moles of NH 4 Cl. g = n * mm = 0.09 moles * 53.5 grams/mole = 4.8 grams of the salt are required.

Example 1.
What is the pH of a C a dissociation constant is K a ?
M acid solution whose acid Solution: Let HA represent the weak acid, and assume x M of it is ionized. Then, the ionization and equilibrium concentration is : HA

H + A C a -x x x K a x 2 = ----- C a -x x 2 + K a x - C a K a = 0 -K a + (K a 2 + 4 C a K a ) 1/2 x = -------------------------- 2 pH = -log(x)

Example 2. Let us make a buffer solution by mixing V a NaA.
mL of acid HA and V s mL of its salt let us assume both the acid and the salt solutions have the same concentration C M.
What is the pH of the so-prepared buffer solution ?
Solution: After mixing, the concentrations C a and C s of the acid HA and its salt NaA respectively are C a = C V a / (V a +V s ) and C s = C V s / (V a +V s ) Assume x M of the acid is ionized. Then, the ionization and equilibrium of the acid is shown below, but the salt is completely dissociated.
HA = H + + A C a -x x x NaA = Na + C s + A C s Common ion [A ] = x+C s K a = x(x+C s ) ------- C a -x x 2 + (K a +C s )x - C a K a = 0 -(K a +C s ) + ((K a +C s ) 2 + 4 C a K a ) 1/2 x = ---------------------------------------- 2 pH = -log(x)

Example 3 Plot the titration curve when a 10.00 mL sample of 1.00 M weak acid HA (K NaOH. a = 1.0e-5) is titrated with 1.00 M Solution A. Because the concentration is high, we use the approximation [H + ] = (C a K a ) 1/2 therefore, pH = 2.500 = 0.00316
B. When 0.1 mL NaOH is added, the concentration of salt (C C s C a s ), and concentration of acid C = 0.1*1.0 M / 10.1 = 0.0099 M = 9.9*1.0 M / 10.1 = 0.98
a are:

HA = H + + A C a -x x x [A ] = x + 0.0099 (= C s ) K a = x (x + 0.0099) ------------------ 0.98 - x x 2 + 0.0099 x = 9.8e-6 - 1e-5 x x 2 + (0.0099 + 1e5) x - 9.8e-6 = 0 = 1e-5 x = (-0.0099 + (0.0099
2 + 4*0.98*1e-5) 1/2 ) / 2 = 0.000907, pH = 3.042 .
Note the sharp increase in pH when 0.1 mL of basic solution is added to the solution.

C. When 1.0 mL NaOH is added, concentration of salt, C s C a = 1.0*1.0 M / 11.0 = 0.0901 M = 9.0*1.0 M / 11.0 = 0.818
HA = H+ + A C a -x x x + 0.0901 K a x (x + 0.0901) = ----------------- 0.818 - x = 1e-5 x = (-0.0901 + (0.0901
2 + 4*0.818*1e-5) 1/2 ) / 2 = 0.0000907 (any approximation to be made?) pH = 4.042
Using the Henderson Hasselbalch equation yields pH = pK a 0.0901(=[A ]) + log (-------------------) = 5 - 0.958 = 4.042 (same result) 0.818(=[HA])

D. When 5.0 mL NaOH is added, concentration of salt, C s C a = 5.0*1.0 M / 15.0 = 0.333 M = 5.0*1.0 M / 15.0 = 0.333 M HA = H+ + A C a -x x x+0.3333
K a = x (x + 0.333) ------------------------- = 1.0e-5 0.333 - x x 2 + (0.333-1e-5)x - 0.333*1e-5 = 0 x = (-0.333 + (0.333
2 + 4*3.33e-6) 1/2 ) / 2 = 0.0000010
pH = 5.000
Note: Using the Henderson Hasselbalch equation yields the same result 0.333(=[A ] pH = pK a + log (---------------------) = 5 + 0.000 = 5.000 0.333(=[HA])

E.
When 10.0 mL NaOH is added, the concentration of salt, Cs = 10.0*1.0 M / 20.0 = 0.500 M and Ca = 0.0*1.0 M / 20.0 = 0.000.At the equivalence point, the solution contains 0.500 M of the salt NaA, and the following equilibrium must be considered: A + H 2 O = HA + OH C s -x x x K b [HA] [OH ] [H + = --------------- --- [A ] [H + ] ] K w 1e-14 x 2 = --- = ------ = 1e-9 = ----- K a 1e-5 C s -x x = (0.500*1.0e-9) 1/2 = 2.26e-5 pOH = -log x = 2.651; pH = 14 - 2.651 = 11.349
Note: The calculation here illustrates hydrolysis of the basic salt NaA. Follow – up: Sketch the titration curve based on the estimates given in this example, and notice the points made along the way. What happens when 1 ml of the NaOH is added beyond the equivalence point ?

Example of an ampholyte - molecule with both acidic and basic groups glycine: pH 1 pH 6 pH 14 NH 3 + – CH 2 – COOH net charge +1 NH 3 + – CH 2 – COO – net charge 0 NH 2 – CH 2 – COO – net charge –1 zwitterion pKa values carboxylate group 2.3
amino group 9.6
can serve as good buffer in 2 different pH ranges. Use glycine to define an important property isoelectric point (pI) - pH at which an ampholyte or polyampholyte has a net charge of zero. for glycine, pI is where: [NH 3 + – CH 2 – COOH] = [NH 2 – CH 2 – COO – ] can calculate pI by applying HH to both ionizing groups and summing (see text) yields: pI = {pK COOH + pK NH 3+ } / 2 = {2.3 + 9.6} / 2 = 5.95
pI is the simple average for two ionizable groups

polyampholytes are molecules that have more than 2 ionizable groups lysine NH 3 + - C- (CH 2 ) 4 - NH 3 +  COOH • • • • titration of lysine shows 3 pKa’s: pH<2, exists in above form first pKa = 2.18, loss of carboxyl proton second at pH = 8.9
third at pH = 10.28
need model compounds to decide which amino group loses a proton first to determine pI experimentally use electrophoresis (see end of Chapter 2)
1.
2.
Gel electrophoresis-electric field is applied to solution of ions, positively charged ions migrate to cathode and negatively charged to anode, at it’s pI an ampholyte does not move because net charge = zero Isoelectric focusing own isoelectric point charged species move through a pH gradient, each resting at it’s Macromolecules with multiples of either only negatively or only positively charged groups are called polyelectrolytes polylysine is a weak polyelectrolyte - pKa of each group influenced by ionization state of other groups

Solubility of macroions (polyelectrolytes and polyampholytes, including nucleic acids and proteins) depends on pH. • • For polyampholytes: high or low pH leads to greater solubility (due to precipitation occurs due to: – or + charges on proteins, respectively) At the isoelectric pH although net charge is zero, there are + and – charges and • • • - charge-charge intermolecular interaction - van der Waals interaction to minimize the electrostatic interaction, small ions (salts) are added to serve as counterions, they screen the macroions from one another
Ionic Strength = I = ½

(M i Z i 2 )
(sum over all small ions) M is molarity,Z is charge Consider the following 2 processes that can take place for protein solutions:
1.
2.
Salting in: increasing ionic strength up to a point (relatively low I), proteins go into solution 2. Salting out: at high salt, water that would normally solvate the protein goes to solvate the ions and protein solubility decreases.
Most experiments use buffers with NaCl or KCl