Definition – solutions which are resistant to changes in their pH when moderate amounts of acid or base are added to them.

An example of a nonbuffered solution is pure H₂O.

Example: adding 10 mL of 0.100 M HCl to 1.00 L of pure water changes the pH from 7 to about 3.

Typical Buffer System – a mixture of a weak acid and its salt or a weak base and its salt (depending on whether we want the resultant buffer pH to be above or below 7).

Consider a buffer consisting of HC₂H₃O₂ and NaC₂H₃O₂, acetic acid and sodium acetate.

HC₂H₃O₂ « H⁺ + C₂H₃O₂^-

For any weak acid HB « H⁺ + B^-

K_a = [H⁺] [B^-] Solving for [H⁺]
           [HB]

You should solve most of these problems as a typical equilibrium-K_a problem. I can never remember these above or what we will later call the Henderson-Hasselblach Equation. You should probably solve these by using the method as the next problem shows.

Calculate the pH of a buffer solution that is simultaneously 0.100 M HC₂H₃O₂ and 0.100 M NaC₂H₃O₂. [You need to look up the K_a for acetic acid.] A buffer like this is referred to as a "symmetrical" buffer because the concentrations of the two ingredients are the same. Solve it as a regular K_a problem.

HC₂H₃O₂ « H⁺ + C₂H₃O₂^-

pH = -Log[1.8 x 10^-5] = 4.74

1. Mix a weak acid (or base) and its salt. In choosing an appropriate acid or base system, you consider what the desired pH of the buffer is supposed to be. A symmetrical buffer will buffer at a pH = pK_a of the weak acid used (or pOH = pK_b of the weak base used). THEREFORE, you select an acid or base that has a pK_a (or a pK_b) close to the pH (or pOH) you want. You can vary the relative amounts of the two components of the buffer to get the actual pH you want.
2. Partially neutralize a solution of a weak acid by using a strong base. This works because the acid molecules that are neutralized are converted to ions, e.g.,

HC₂H₃O₂ + OH^- ® C₂H₃O₂^- + H₂O

If only a part of the acid molecules is neutralized, we create a solution containing a weak acid and its salt.

You can accomplish #2 by ½ Neutralizing a weak acid.

HB « H⁺ + B^-

[H⁺] = K_a x [HB] / [B^-]

If you ½ neutralize the Acid [HB]= [B^-] & [H⁺] = K_a

3. Add a strong acid to a solution of the salt of a weak acid using fewer moles of the strong acid than you have of the salt of the weak acid.

The Reation is: H⁺ + C₂H₃O₂^- ® HC₂H₃O₂

250.0 mL of 0.500 M NaOH are added to 350.0 mL of 0.900 M HC₂H₃O₂. Calculate the pH of the resultant buffer.

HC₂H₃O₂ ® H⁺ + C₂H₃O₂^-                 K_a= 1.8x10^-5

I We want to get these – we want INT. # moles of HC₂H₃O₂ added.

C

E

(.3500L) (.900 Mole / L) = .0315 Mole HC₂H₃O₂ # moles of OH^- added.

(.3500L) (.500 Mole / L) = .0125 Mole of OH^-

SO

# Moles C₂H₃O₂^-             Formed = .0215 Moles C₂H₃O₂^-

# Moles HC₂H₃O₂         Remaining .0315 - .0215 = .190 Moles

Total Volume = 250.0 + 350.0 = 600.0 ML

K_a = 1.8 x 10^-5 = (x) (.125/.6 + x) / (.190 /.6 –x) Assume x is small.

HB + OH^- ® H₂O + B^-

[ALL SPECIES ARE (aq).]

B^- + H⁺ ® HB

As a result of these reactions, H⁺ and OH^- ions are consumed, so the pH does not change dramatically.

Moles of HC₂H₃O₂ and C₂H₃O₂^- before

(1.00L) (.100 Mol/L) = .100 Moles each

Moles of H+ added

(.100 L) (.100 Mol/L) =.0100 moles of H⁺

Reaction that takes place

After this reaction

# Moles C₂H₃O₂^- = .100 - .0100 = .090 Mole

# Moles HC₂H₃O₂ = .100 + .0100 = .0110 Mole

HC₂H₃O₂ « H⁺ + C₂H₃O₂^-

Total Vol = 1.10 L

k_a = [H⁺][C₂H₃O₂^-] = (x)(0.090/1.10L + x)
         [HC₂H₃O₂]        (0.110/1.10L – x)

Drop x because it is small

1.8 x 10^-5 = (x)(0.090/1.10) = 2.2 x 10^-5 M
                    (0.110/1.10)

pH = +4.66

Therefore not much of a change

There is an equation that many teachers like to utilize when working with buffer solution problems called the Henderson-Hasselbach equation [see p. 512], derived from the equilibrium constant expression, K_a or K_b.

I fail to understand why anyone really wants to use this equation. It just seems like one more thing for students to memorize and it would seem relatively easy for students to mix up the equation and get it "upside down." I have always wondered how Henderson and Hasselbalch (I assume they are two different individuals) managed to get their names in our textbooks for something that looks so trivial. Anyway, here is how the equation is derived. On the second thought forget it.

Prepare 500. mL (3 sig figs!) of a buffer solution that has a pH of 7.00 by using only 0.500 M hypochlorous acid, HClO, and 0.350 M sodium hypochlorite, NaClO. The K_a for HClO = 2.95 x 10^-8.

Since the pH is 7.00, the [H⁺] = 10^-7. We want to know the volumes of HClO and NaClO needed.

Let x = volume in liters of 0.500 M HClO required. Then (.500-x) = volume in liters of 0.350 M NaClO required.

 [ClO-]   = (.500-x) (.350)/0.500 L = .295
[HClO]        (x) (.500)/0.500L

Solving for x, x = 0.352 L so we will use 352 mL of 0.500 M HClO and (500 - 352) = 148 mL of 0.350 M NaClO.

This refers to the amount of acid or base that can be added to a buffer solution before its pH changes "appreciably." It can be demonstrated that the most effective buffer exists when the concentrations of both the weak acid and its salt (or weak base and its salt) are high and are equal to each other, a symmetrical buffer. A buffer, therefore, has its maximum capacity when pH = pK_a (or pOH = pK_b). Whenever the ratio of salt to weak acid or base becomes either less than 0.10 or more than 10.0, the buffer begins to lose its effectiveness. Thus, a buffer is most effective in a range about one pH unit on either side of the pK_a (or pK_b) of the weak acid or base present in the buffer system.

It is absolutely essential that the pH of human blood and intercellular fluids remain fairly constant, because the functioning of many enzymes that operate in human biochemistry is often sharply pH dependent.

The pH of human blood is 7.4. Severe illness or death results if the pH varies even a few pH units from this value for any sustained period of time.

Acidosis (low blood pH) can result from heart failure, kidney failure, diabetes, persistent diarrhea, or a long-term high protein diet. Prolonged, intense exercise can cause a temporary condition of acidosis.

Alkalosis (high blood pH) can be caused by severe vomiting, hyperventilation, or exposure to high altitudes.

Studies performed on climbers who reached the summit of Mount Everest (29,028 ft) without supplemental oxygen had arterial blood with a pH of 7.7-7.8. Extreme hyperventilation required to compensate for the low partial pressure of O₂ (about 43 mm Hg) caused this condition.

Blood is a very complex buffer. The major buffering system involves the carbonic acid-bicarbonate ion system (actually CO₂ and HCO₃^-). Interestingly enough, the ratio of HCO₃^- to H₂CO₃ is about 10:1, which would seem to place this buffer outside the range of its maximum buffering capacity. The situation is rather complex, but this seems useful because:

Indicators are not "magical" substances. They are simply weak acids (usually) which happen to have one additional and useful property--the color of the molecular (acid) form of the indicator, HIn, is a different color from the ionic (basic) form of the indicator, In^-.

Since they are weak acids:

This shows that the 2 forms of the indicator depend upon the [H+] in the solution. Since the [H+] changes by several powers of 10 near the equivalence point of a titration, the ratio of the two forms of the indicator changes by this same factor. That is why the color changes so rapidly and dramatically.

When an indicator is just ready to change color, [HIn] = [In-].

This means that

[H+] [In-] = K_a     so [H+] = K_a         and pH = pK_a
   [HIn]

Ex. Consider the titration of 25.00 mL of 0.1000 M HCl by 0.1000 M NaOH. Calculate the pH after 10.00 mL of base has been added.

          Initial # Mol H⁺      
(0.02500L)(0.1000 Mol/L) = 0.002500 Mol H⁺

        Mol OH^- Added       
(0.01000L)(0.1000 Mol/L) = 0.001000 Mol OH^-

#Mol H⁺ remaining = 0.002500-0.00100
                              = 0.001500 Mol H⁺

pH = -log(0.0426) = 1.3680

DATA for pH of the solution vs volume of NaOH added:

Titration curve for a strong acid by a strong base--25.00 mL of 0.1000 M HCl by 0.1000 M NaOH.

VERY IMPORTANT: Read p. 521, top 3 paragraphs. See Fig. 15.8. Read #1 and #2 under Fig. 15.8.

Consider the titration of 25.00 mL of 0.1000 M HC₂H₃O₂ with 0.1000 M NaOH. Let’s look at sample calculations of pH of the solution at various points during the titration:

1. For the original solution find pH before any base added. What is the chemistry you need to solve this?

x = 1.3 x 10^-3                 So     pH = 2.87

b) At the equivalence point of the titration: at this point, the neutralization reaction has "gone to completion," and what we have is essentially a solution of sodium acetate, NaC₂H₃O₂. Since the acetate ion is the conjugate base of a weak acid, it will hydrolyze according to the following reaction:

C₂H₃O₂^- + H₂O « HC₂H₃O₂ + OH^-

K_a of acetic acid = 1.8 x 10^-5

AT THE END POINT

Vol Total = 50.00 mL

5.6 x 10^-10 =      x²              à       x = 5.3 x 10^-6 M = [OH^-]
                  0.0500 – x     small x

pOH = -log[5.3x10^-6M] = 5.28

pH = 8.72

Even though we are just past the equivalence point, we can make the assumption that virtually all of the OH^- in the solution will come from the excess, unneutralized base. The amount coming form the hydrolysis from the acetate ion can be neglected.

Since the acid and base were of the same concentration, the original 25.00 mL of acid required 25.00 mL of base to be neutralised. Therefore, there are 1.00 mL of base that are in excess, and [note total volume is now 51 mL]

#Moles OH^- left over = (0.00100L)(0.1000 Mole/L)

                            = 1.00 x 10^-4 Mole OH^-

pOH = -Log 1.96 x 10^-3 = 2.708

pH = 14 – 2.71 = 11.292

1. The initial pH is higher than in the titration of a strong acid because the weak acid is only partially ionized.

2. There is a rather sharp increase in pH at the start of the titration.

3. Over a long section of the curve preceding the equivalence point the pH changes quite gradually. Solutions for this portion of the curve are buffer solutions.

4. At the point of half-neutralization, pH = pK_a. At the half-neutralization point, [HA] = [A^-].

5. The pH at the equivalence point is greater than 7. The anion of the salt hydrolyses because it is the conjugate base of a weak acid.

6. Beyond the equivalence point the titration curve is identical to that of a strong acid by a strong base. In this portion of the titration the pH is established by the concentration of unreacted OH-.

7. The steep portion of the titration curve at the equivalence point is over a relatively short pH range (for example, from about pH 7 to 10).

8. The selection of indicators available for the titration is more limited than in a strong acid-strong base titration.

Therefore, for a weak acid-strong base titration, you need an indicator that changes color above pH 7. This means the pKa for the indicator is >7. The opposite for a strong acid & weak base.

Consider 50.00 mL of 1.000 M NH₃ titrated with 1.000 M HCl looking at 3 stages in the titration.

a) before any hydrochloric acid is added:

b) after 25.00 mL of the HCl has been added:

c) after 50.00 mL of acid has been added, that is, at the equivalence point:

d) after 51.00 mL of acid has been added:

Consider the titration of 10.00 mL of 0.1000 M H₃PO₄ with 0.1000 M NaOH.

Essentially three separate neutralizations take place in successive order. The equations for these neutralizations are respectively

H₂PO₄^- + OH^- à H₂O + HPO₄^2-

HPO₄^2-+ OH^- à H₂O + PO₄^3-

Corresponding to these three distinctive stages, we would expect there to be three equivalence points. For every mole of H₃PO₄, 1 mole of NaOH is required to reach the first equivalence point. At this first equivalence point, the solution is essentially just a solution of NaH₂PO₄. This solution should be somewhat acidic, because the acid ionization constant of H₂PO₄^- (Ka = 6.34 x 10^-8) is greater than the base ionization (hydrolysis) constant, (K_b = 1.41 x 10^-12).

H₂PO₄^- + H₂O « H₃O⁺ + HPO₄^2-         Ka = 6.34 x 10^-8

H₂PO₄^- + H₂O « H₃PO₄ + OH^-                 K_b = 1.41 x 10^-12

To reach the second equivalence point, 1 more mole of NaOH will be required. Now we essentially have a solution of Na₂HPO₄. This solution will be somewhat basic, because the base ionization (hydrolysis) constant of HPO₄^2- is greater than the acid ionization constant of HPO₄^2-.

HPO₄^2- + H₂O « OH^- + H₂PO₄^-                K_b = 1.58 x 10^-7

Although you anticipate a third equivalence point, none is present. This is because at the third equivalence point you will have what essentially is a solution of Na₃PO₄. Phosphate ion is such a strong Bronsted-Lowry base that it hydrolyses extensively. In fact the pH may, depending on the concentration of the solution, be higher than 13, which is the pH of the 0.1000 M NaOH that we are using in the titration. Thus, there is no way that we can show a rapid increase in pH as add the NaOH, so no third equivalence point can be detected.

A. Expression for K_sp: AgCl(s) « Ag⁺_(aq) + Cl^-_(aq)

                            K_sp = 1.6 x 10^-10 = [Ag⁺] [Cl^-]

                            PbCl_2(s) « Pb²⁺_(aq) + 2 Cl^-_(aq)

                            K_sp = 1.7 x 10^-5 = [Pb²⁺] [Cl^-]²

B. Uses for K_sp

1. Calculation of concentration of one ion, knowing that of another. What is the [Pb²⁺] in a solution in equilibrium with PbCl₂ if [Cl^-] = 0.020 M?

(k_sp = 1.7 x 10^-5) if [Cl^-] = 0.020 M?

2. Determination of whether precipitate will form. If Q (the ion product set up like the Ksp expression) is less than K_sp, no precipitate will form (equilibrium not established). If Q > K_sp, precipitate forms until Q becomes equal to K_sp.

Suppose enough Ag⁺ is added to a solution 0.001 M in CrO₄^2- ion to make the concentration of Ag⁺ equal to 0.001 M. Assuming the concentration of chromate remains unchanged, will silver chromate precipitate? The K_sp = 2 x 10^-12.

Ag₂CrO₄ « 2Ag⁺ + CrO₄^2-

Q = (oring. Conc. Ag⁺)² x (orig. conc. CrO₄^2-)

3. Determination of solubility READ p. 536, bottom

1. PbCl₂(s) in pure water

K_sp = [Pb²⁺_(aq)] [Cl^-_(aq)] ²       next plug in!

4s³ = K_sp = 1.7 x 10^-5

So s = 1.6 x 10^-2M

b) In solution containing a common ion: example solubility of PbCl₂(s) in 0.100 M HCl. READ p. 538, bottom.

Or solving this using K_sp s(0.10)² = 1.7 x 10^-5
                                                              s= 1.7 x 10^-3M

WHAT IS THE SOLUBILITY OF X₃Y₂ IN TERMS OF K_sp?

X₃Y₂ « 3X²⁺ + 2Y^-3                     K_sp = [X²⁺]³[Y^3-]²

Let 3s = [X²⁺]                               2s = [Y^3-] so

K_sp = (3s)³(2s)² = 27s³ x 4s² = 108s⁵

Or s = (K_sp/108)^1/5

1) List the major species in the solution.

2) Look for reactions that can be assumed to go to completion, for example, a strong acid dissociating or H+ reacting with OH-.

3) For a Rx that can be assumed to go to completion:

        a) determine the concentration of the products.

        b) write down the major species in solution after the Rx.

4) Look at each major component of the solution and decide if it is an acid or a base.

5) Pick the equilibrium that will control the pH. Use known values of the dissociation constants (K_a and K_b) for the various species to help decide on the dominant equilibrium. Which has a larger value?

a) Write the equation for the Rx and the equilibrium expression.

b) Compare the initial concentrations (assuming the dominant equilibrium has not yet occurred, that is, no acid dissociation, etc.)

           c) Define x.

           d) Compute the equilibrium concentrations in terms of x.

 TABLES FOR "INIT, CHANGE, AND EQ’M" HELP CLARIFY THIS.

e) Substitute the concentrations into the equilibrium expression and solve for x.

            f) Check the validity of the approximation.

            g) Calculate the pH and other concentrations as required.

Copied from pp 658-659 of Steven S. Zumdahl, Chemistry, 2nd edition, 1989.