# Acids and Bases, cont. Titration Curves, cont.

#### Titration of a weak acid with a strong base: (lecture slide, Zumdahl Fig 15.3, p 722)

The curve below shows the titration of 1M acetic acid with 1M NaOH. In this case the pH is no longer a simple function of the concentration of acid, since the acid only partially dissociates in water. It does however react completely with any added strong base until the acid has been consumed, so adding base has little effect on pH initially. Let's look at the titration curve of 1M acetic acid with 1M NaOH. The reaction is:

HC2H3O2 + OH- C2H3O2- + H2O

• The curve starts at about pH 3 for the dilute acetic acid. At this point there is approximately 99% HC2H3O2 in the solution.

• The curve first rises fairly quickly, as the free protons from the dissociated acetic acid are consumed.

• The curve then rises in a nearly linear fashion (red tangent line superposed on curve), with each hydroxide added consuming an acetic acid molecule. converting it to acetate ion. This region of the curve, where it is approximately linear, is known as the Buffer Region.
• When the volume of sodium hydroxide added equals the initial volume of acid (equal moles of each have been added), the amount of acid becomes equal to the amount of salt: there is 50% HC2H3O2 and 50% C2H3O2-.
• Note that by the Henderson-Hasselbalch equation we now have pH = pKa since log 50/50 = log 1 = 0.
• If we now draw a line from this point on the curve parallel to the x-axis, it will intersect the y-axis where pH = pKa.
• As we approach the point where equal amounts of base and acid have been added we see an increasing slope, until it reverses slope (there is an inflection) becoming almost vertical. This inflection point is the equivalence point, that is the point where equal quantities (moles) of acid and base have been added to the solution.
• Finally, as more base is added, the curve levels off again as we approach the pH of the pure base solution. We can calculate the pH at any point in the process of a titration.

Example: 20.0 mL of 0.10 M NaOH is added to 50.0 mL of 0.10 M HCHO2. What is the pH of the resulting solution? Ka = 1.8 x 10-4, pKa = 3.74

 HCHO2 + OH- H2O + CHO2- Before rxn (0.050)(0.10)/0.070 (0.020)(0.10)/0.070 0 at equilibrium (0.030)(0.010)/0.070 0 (0.020)(0.10)/0.070

Notice that we now have a buffer (both an acid and its conjugate base present in significant amounts)! Can use Henderson-Hasselbalch equation:

pH = pKa + log (0.020)(0.010)/0.070 / (0.030)(0.10)/0.070

= pKa + log 0.020/0.030

pH = 3.74 + (- 0.176) = 3.56

Note that the pH is < pKa, as expected since more acid is present than salt!

For a deeper understanding of this titration curve you may like to try the Excel™ based titration exercise and questions I have prepared.

The titration of a weak base with a strong acid looks very much the same except it is mirrored around a horizontal axis. (Zumdahl, Fig 15.5, p 728)

Note that whenever you have a strong acid or base involved in a titration or acid-base reaction the reaction will go until ALL of one or both of the reactants is consumed!

Example: mix 20.0 mL 0.10M NaOH with 15 mL 0.10 M H2SO4 what happens?

• Initially have (0.0200L)(0.10mol/L) = 0.0020mol OH- and (0.015L)(0.10mol/L)(2eq/mol) = 0.0030mol H+ available for reaction.
• After reaction ALL of sodium hydroxide is consumed, ALL (0.0015mol) of first dissociable protons and (0.0005mol of HSO4-) consumed, but 0.0005 mol remains (have 0.0005mol HSO4- and 0.0005mol SO42-)

# Indicators

Frequently it is convenient to use an indicator to show when an equivalence point has been reached. What we are attempting here is to see when exactly equal amounts of acid and base have been added to the solution. Usually, however, what is actually indicated is that a slight excess of the titrant (base in our examples) has been added. An indicator is then a substance which changes color (most commonly) when it reacts with a base. Thus indicators are weak acids which absorb light differently in their protonated acidic form than they do in their ionized form. (Zumdahl Fig 15.8, p 732)

This means that indicators have titration curves just like the curve for acetic acid, but with different pKa's. In order to be an effective indicator, with little error, its "buffer region" must occur in the nearly vertical region of the titration curve of the substance we are titrating (e. g. pH 7 - 11 in the titration curve above. That way a complete color change will take place with very little addition of base past the equivalence point and thus very little error. (Zumdahl Figures 15.9 and 15.10, p 733) For example, phenolphthalein has a color-change (buffer region) from 8.3(colorless)-10.0(fushia).

Note also that since the indicator is itself a weak base we must add only a tiny amount, or its own titration will add a significant error. Indicators must therefore be intensely colored.

Finally, for a particularly accurate titration, one can titrate a blank - a volume of water with the same number of drops of indicator as the sample, and subtract the volume of titrant needed for this blank from the amount needed in the actual titration.

# Hydrolysis (Acid/Base Behavior of Salts)

When a weak acid is titrated with a strong base it is found that the equivalence point always occurs at a pH above neutrality (pH > 7). Similarly, when a weak base is titrated with a strong acid the equivalence point always occurs at a pH below neutrality (pH < 7). Why? Essentially it is a consequence of the the nature of acids as seen in the Brønsted acid description: when the acid is titrated a weak base, the conjugate base is formed. The conjugate base then competes with hydroxide ion for hydrogen ion, leading to a slightly higher concentration of hydroxide ion as the equilibrium is shifted:

A- + H2O OH- + HA

This process is sometimes referred to as hydrolysis, which is the term we will also use. (Notice that Zumdahl discusses this phenomena in section 14.8 Acid-Base Properties of Salts pp 671-77.)

Note that the ion of a strong acid, such as HCl, will not affect pH since the strong anion (e.g chloride ion) has no tendency to react with water. Similarly the ion of a strong base such as NaOH will not affect pH since the strong cation (e.g. Na ion) has no tendency to react with water either.

So how is the pH affected by the presence of the salts of weak acids? Let's look at our favorite acid, acetic acid reacting with NaOH:

HC2H3O2 + OH- C2H3O2- + H2O

The resulting acetate ion can now react with water:

C2H3O2- + H2O HC2H3O2 + OH-

This reaction can be written as the sum of the association of acetic acid and the dissociation of water:

C2H3O2- + H+ HC2H3O2

H2O H+ + OH-

Notice that the the equilibrium constant for the association of acetic acid is the inverse of the dissociation constant (the reaction is backwards, inverting the equilibrium expression): K = 1/Ka

The overall equilibrium constant is then the product of the equilibrium constants for the two reactions, Kh:

Kh = (Kw)(1/Ka)

Kh is in fact Kb for the conjugate base. (Zumdahl use Kb in his discussion.) Kb, conj = Kw/Ka

A similar treatment is seen for weak bases. Using ammonia as an example:

NH3 + H2O NH4+ + OH-

When an ammonium salt is dissolved in water the ammonium ion reacts with the water:

NH4+ + H2O H3O+ + NH3

This time Kh is Ka for the conjugate acid, NH4+, so

Kh = Ka = [H+] [NH3] / [NH4+] = Kw/Kb

Note that the weaker the acid or base, the greater the hydrolysis of its conjugate ion (the more the pH deviates from neutrality)!

Let's look at some examples qualitatively. As we look at them let's name the salts; write the formulae and name the acids corresponding to the ions, and write the reactions for the various ions with water as the basis for our predictions.

 Salt pH KCN strong base, weak acid salt K+ + H2O NR CN- + H2O HCN + OH- >7 KC2H3O2 strong base, weak acid salt K+ + H2O NR C2H3O2- + H2O HC2H3O2 + OH- >7 NH4Cl weak base, strong acid salt NH4+ + H2O NH3 + H3O+ Cl- + H2O NR <7 NH4 NO3 weak base, strong acid salt NH4+ + H2O NH3 + H3O+ NO3- + H2O NR <7 NH4CN NH3 stronger than HCN NH4+ + H2O NH3 + H3O+ CN- + H2O HCN + OH- >7 NH4C2H3O2 pKb NH3 = pKa HC2H3O2  NH4+ + H2O NH3 + H3O+ C2H3O2- + H2O HC2H3O2 + OH- 7

### NEXT

 Syllabus / Schedule C109 Home © R A Paselk