Salt Solutions.

In the equilibrium:

HA + H2O H3O+ + A-

we note that the reverse reaction has A- acting as a base and taking a proton from H3O+.

If A- is a base, it can take a proton from neutral water:

A- + H2O HA + OH-

To what extent does this reaction proceed?

What is Kb for this reaction?

It turns out that the K values for conjugates are related.

To see this relationship, add the above two equations:

 HA + H2O H3O+ + A- A- + H2O HA + OH- Ka Kb(A-) 2 H2O H3O+ + OH- Kw

Since adding reactions means multiplying Ks, we get:

Ka(HA) * Kb(A-) = Kw = 1 x 10-14

or (taking logs)

pKa + pKb = pKw = 14 for conjugates.

Deciding whether a conjugate will react with water to any significant extent, or not.

1. For conjugate bases of HA.

pKa(HA) + pKb(A-) = 14.

or

pKb(A-) = 14 - pKa(HA)

Rules:

If pKa or pKb > 14, the acid or base will not react significantly with water.

If pKa or pKb < 0, the acid or base will completely react with water and the equilibrium can be ignored.

The consequence of these:

If pKa(HA) < 0, then pKb(A-) > 14 and the anion's reaction with water can be ignored.

If pKa(HA) > 0 but < 14, then the pKb(A-) < 14 but > 0 and the equilibrium calculation must be used.

2. For conjugate acids of B:

If pKb(B) < 0, then pKa(BH+) > 14 and the cation's reaction with water can be ignored.

If pKb(B) > 0 but < 14, the the pKa(BH+) < 14 but > 0 and the equilibrium calculation must be used.

Hydrolysis of Salts.

The reaction of a cation or an anion of a salt with water is termed salt hydrolysis.

An approximate way of looking at whether hydrolysis will be important or not is the following table:

 substance conjugate strong acid hydrolysis insignificant weak acid weak base weak base weak acid strong base hydrolysis insignificant

We will deal with salt solutions that contain no more than one hydrolysable ion.

Salts of strong acids and strong bases contain no hydrolysable ions and so are always neutral (pH = 7), no calculation is necessary.

Salts of strong acids and weak bases are always acidic, a calculation is necessary.

Salts of weak acids and strong bases are always basic, a calculation is necessary.

Salts of weak acids and weak bases are not discussed.

The special case of metal hydroxides.

The metal hydroxide salts are a little different in that when they dissociate they produce OH-. Since these compounds can be considered as salts, they are fully dissociated, and that is how the pH of a solution containing one of these can be determined. Note that since the salt is fully dissociated, there is no tendency for the cation (the metal ion) to react with water.

However, for future reference:

This is not strictly true, since metal ions can act as Lewis acids with water as the Lewis base to provide hydrated metal complex ions such as Cu(H2O)42+. It is possible for these ions to acts as Br�nsted acids:

Fe(H2O)63+ + H2O Fe(H2O)5(OH)2+ + H3O+.

We will ignore this possibility.

Example calculation of salt pH:

Calculate the pH of a 0.100 M solution of sodium acetate.

pKa (CH3COOH) = 4.75.

In calculations involving salts:

1. Always assume that the salt is a strong electrolyte and so is fully dissociated into ions in aqueous solution:

CH3COO- Na+(aq) CH3COO-(aq) + Na+(aq)

2. Decide which ion will react with water, and which will not. We will not deal with the situation where both ions react with water.

In this case, the Na+ is the cation in NaOH and so will not react. The CH3COO- is the conjugate base of the weak acid CH3COOH and so will react:

 CH3COO- + H2O CH3COOH + OH- I (M) 0.100 0 0 C (-1)x (+1)x (+1)x E 0.100 - x x x

Kb = a(CH3COOH) * a(OH-) / a(CH3COO-)

Kb = x * x / (0.100 - x)

To solve we need to know one of the unknowns. Since pKa for acetic acid is given, we can calculate pKb for acetate ion:

pKb(CH3COO-) = 14 - pKa(CH3COOH)

= 9.25

and

Kb(CH3COO-) = 5.62 x 10-10.

Substituting:

Kb = a(CH3COOH) * a(OH-) / a(CH3COO-)

= x * x / (0.100 - x)

5.62 x 10-10 = x * x / (0.100 - x)

Using the approximation that x is small:

x = 7.5 x 10-6.

Check approximation: 7.5 x 10-6 * 100% / 0.1 = 7.5 x 10-3%

Approximation is valid.

x = a(OH-), so - log x = pOH = 5.1

Thus the pH of this salt solution is 8.9.

Example number 2:

Calculate the pH of a solution of 10.0 g of trimethyl ammonium chloride, (CH3)3NH+ Cl-, dissolved in 1.00 L water. Assume no change in liquid volume on dissolution.

pKb of trimethylamine, (CH3)3N is 4.19.

Step A: Analysis:

1. The salt is fully dissociated into ions:

(CH3)3NH+ Cl- (CH3)3NH+(aq) + Cl-(aq).

2. Cl- comes from the strong acid HCl and so is not hydrolysed.

3. a( (CH3)3NH+) = 10.0g (CH3)3NH+ Cl- /95.5 g/mol * 1/1.00L

= 0.105 M

3. Hydrolysis of the trimethylammonium ion:

 (CH3)3NH+ + H2O (CH3)3N + H3O+ I (M) 0.105 0 0 C (-1)x (+1)x (+1)x E 0.105 - x x x

Ka = a( (CH3)3N) * a(H3O+) / a( (CH3)3NH+)

= x * x / (0.105 - x) = 1.55 x 10-10

Step B: Brainstorm (trivial, find x). Step C: Calculate.

Assume x small: x = 4.03 x 10-6 (so approximation is valid)

pH = - log x = 5.39.

The pH of a solution of 10.0 g of trimethyl ammonium chloride, (CH3)3NH+ Cl-, dissolved in 1.00 L water is 5.39.

Titrations.

Salt solutions are important in titrations. A titration is the addition of an acid or a base to a base or acid in such a way that quantitative relations may be obtained.

When an acid and a base react with each other, the products are usually a salt and water. Consequently, the product of a titration is a salt, and the pH of the solution at the end of a titration is the pH of the salt solution. As we have just seen, this may or may not be neutral.

 titration between: pH at end of reaction: strong acid + strong base neutral weak acid + strong base basic strong acid + weak base acidic weak acid + weak base ? (do not perform)

To calculate the pH at the end of a titration:

if both are strong, pH = 7 (no calculation)

if one is weak, calculate the activity of the salt solution at the end of the titration, then calculate the pH as for salt solutions.

To calculate the pH of a solution during titration see below.

Finding the end of a titration process.

There are two ways in which the end of a titration can be obtained.

1 Use a pH meter to measure the pH at different points during the titration. The point found in the middle of the S curve is the exact point where the same amount of acid and base have been added.

This exact point is called the equivalence point.

2 Use an indicator to show when the reaction is over.

Indicators.

Acid - base indicators are weak acid or weak base dyes which have different colours for their conjugates.

 HIn + H2O H3O+ + In- colour 1 colour 2

When the indicator reacts, it changes colour:

 HIn + H2O H3O+ + In- colour 1 colour 2

Clearly, if this colour change is to occur so that it indicates that the main reaction is over, two things must be true:

For acids being titrated against a strong base

1. The indicator must react after the main reaction has taken place. This will be the case if the indicator is a weaker acid than the acid in the titration. To be sure that the indicator does not react too soon, the indicator should have a pKa at least 2 units higher than the pKa of the acid being titrated.

2. The colour change must be quick, the indicator must not use much acid to change. To be sure, very little indicator is used.

For bases being titrated against a strong acid

1. The indicator must react after the main reaction has taken place. This will be the case if the indicator is a weaker base than the base in the titration. To be sure that the indicator does not react too soon, the indicator should have a pKb at least 2 units higher than the pKb of the acid being titrated.

2. The colour change must be quick, the indicator must not use much base to change. To be sure, very little indicator is used.

Because the change in colour of the indicator may not indicate exactly the equivalence point in a titration, the point the indicator shows is called the end point in the titration.

Choice of an indicator (so that the end point is close to the equivalence point):

For strong acid - strong base titrations:

any indicator will do (all are weak acids).

For weak acid - strong base titrations:

use indicator with a pKa = 8 or more

for example: phenolphthalein (pKa = 9.4)

For strong acid - weak base titrations:

use an indicator with a conjugate base with pKb = 8 or more

for example: methyl orange (pKa = 3.4,

pKb of the conjugate ion = 10.6)

Indicators: a quantitative look.

A useful indicator changes colour at the pH of the equivalence point so indicating that the main reaction is complete.

Indicators change colour over a range of pH, and to choose an indicator for a particular titration we need to know the pH at the end of the titration (the pH of the salt solution), and the pH range for indicators.

Consider the indicator equilibrium:

 HIn + H2O H3O+ + In- colour 1 colour 2

Ka = a(H3O+) * a(In-) / a(HIn)

in logarithms:

pKa = pH - log {a(In-) / a(HIn)}

At the start of the titration, the indicator will be present entirely as either HIn (in acid) or In- (in base).

Assume that the indicator is in acid solution, and that base is being added.

Further assume that when [HIn] is 10 x [In-], colour 1 will show, and when [In-] is 10 x [HIn], colour 2 will show.

Substitute these values into the pKa equation:

pKa = pH - log {a(In-) / a(HIn)}

colour 1 pKa = pH - log { 1 / 10 }

= pH - (-1)

= pH + 1

colour 2 pKa = pH - log { 10 / 1 }

= pH - 1

The change from colour 1 to colour 2 with these assumptions occurs as the pH of the solution changes from pKa - 1 to pKa + 1

Similarly, in base solution, it can be shown that the colour change occurs as the pH of the solution changes from

pKa + 1 to pKa - 1.

(See table of indicators, p 830 in your text to see how closely this is followed.)

Obtaining the pKa of a weak acid (or pKb of a weak base) from titration data.

The titration curve for an weak acid - base titration obtained from a pH meter is: (For a weak base - acid titration curve, see page your text p827.)

1. The starting point for the curve can be calculated as the pH of the acid solution.

2. The equivalence point can be calculated as the pH of the resulting salt solution.

3. In the flatish part of the titration curve, the pH can be calculated from the ratio of the amount of acid left to salt produced:

from the definition of Ka for an acid:

Ka = a(H3O+) * a(A-) / a(HA)

or in logs: pKa = pH - log { a(A-) / a(HA) }

so pH = pKa + log { a(A-) / a(HA) }

Example: Calculate the pH curve for the titration of 25 mL of 0.100 M acetic acid with 0.100 M sodium hydroxide.

1. At start of reaction, calculate pH of the acetic acid solution:

(pKa(acetic acid) = 4.75)

 CH3COOH + H2O H3O+ + CH3COO- I (M) 0.100 0 0 C (-1)x (+1)x (+1)x E 0.100 - x x x

Ka = {a(H3O+) * a(CH3COO-)} / a(CH3COOH)

1.78 x 10-5 = x * x /(0.100-x)

unknown: pH, calculated from x:

Solution: x = 1.33 x 10-3, pH = 2.87.

2. At end, calculate the pH of the salt solution, CH3COONa:

(Don't forget dilution!)

CH3COO- + H2O CH3COOH + OH-

 CH3COO- + H2O CH3COOH + OH- I (M) 0.100 0 0 C (-1)x (+1)x (+1)x E 0.100 - x x x

Kb = a(CH3COOH) * a(OH-) / a(CH3COO-)

Kb = x * x / (0.100 - x)

5.62 x 10-10 = x * x / (0.0500 - x)

Solution: x = 5.3 x 10-6; pOH = 5.28; pH = 8.72.

3. During the titration:

pH = pKa + log { a(CH3COO-) / a(CH3COOH) }

To calculate the two concentrations:

 CH3COOH + NaOH CH3COO-Na+ + H20 I 2.5 mmol 0 0 C (-1)x (x added) (+1)x End (2.5 - x) mmol 0 x

Ratio of concentrations is:

[CH3COO-] / [CH3COOH] = x mmol / (2.5 - x)mmol

(note the volume of solution is the same for both species and cancels)

and from the above:

pH = pKa + log { a(CH3COO-) / a(CH3COOH) }

= 4.75 + log { x / (2.5 - x)}.

 mL base mmol base log {x/(2.5-x)} pH 3 0.3 -0.87 3.88 5 0.5 -0.60 4.15 10 1 -0.18 4.57 15 1.5 0.18 4.93 20 2 0.60 5.35 22 2.2 0.87 5.62 23 2.3 1.06 5.81 24 2.4 1.38 6.13

Calculation of the pKa for a weak acid (or a weak base).

 HA + NaOH A- Na+ + H20 I a mmol 0 0 C (-1)x (x added) (+1)x End (a - x) mmol 0 x

pKa = pH - log { a(A-) / a(HA) }

pKa = pH - log { x / (a - x)}.

By measuring the pH of a solution part way through a titration, and knowing how much base was added, the pKa of the acid can be found.

Example:

To 25.0 mL of a 0.100 M solution of nitrous acid (HNO2) is added 6.00 mL of 0.15 M sodium hydroxide solution. The pH of the resulting solution is found to be 2.89. Calculate the Ka for nitrous acid.

 CH3COOH + NaOH CH3COO-Na+ + H20 I 2.5 mmol 0 0 C (-1)x (x added) (+1)x End (2.5 - x) mmol 0 x

pKa = pH - log { a(A-) / a(HA) }

pKa = pH - log { x / (a - x)}.

a (initial amount of acid) = 2.5 mmol

x (amount of base added) = 0.900 mmol

= 2.89 - log { 0.9 / (2.5 - 0.9)}

= 2.89 - (-0.25)

= 3.14

From this data, the pKa(HNO2) = 3.14.

An Especially Easy Calculation of pKa (or pKb) from a titration curve.

pKa = pH - log { a(A-) / a(HA) }

pKa = pH - log { x / (a - x)}.

Take the situation exactly half way through the titration. At this point x = � a, so:

pKa = pH - log { � a / (a - � a)}.

pKa = pH - log { 1 }.

pKa = pH

Half way through a titration, the pKa of the weak acid is equal the pH of the solution. Buffer Solutions.

A buffer solution is defined as a solution which resists a change in pH.

Note that it doesn't stop a change in pH completely.

Since a buffer must resist a change in pH when an acid is added, it must contain a base.

Similarly, to resist a change in pH when a base is added, it must contain an acid.

Normally, an acid will react with a base.

The simplest way of creating a buffer is to use an acid-base conjugate pair in equilibrium:

 HA + H2O A- + H3O+ acid base

 B + H2O BH+ + OH- base acid

To make a buffer using a conjugate pair:

dissolve some of the acid and some of the sodium salt of the acid in solution.

or dissolve some of the base and some of the hydrochloride salt of the base in solution.

Making Up a Buffer Solution.

From the definition of Ka for an acid:

Ka = a(H3O+) * a(A-) / a(HA)

or in logs: pKa = pH - log { a(A-) / a(HA) }

1. to be a useful buffer, there must be an adequate amount of both the acid and the base present. The concentration of both the acid and the base must be no less than 0.1 M.

2. so that the pH can be calculated, both acid and base solutions must be dilute. The concentration of both the acid and base must be no more than 1.0 M

Extremes for buffer solutions:

acid to base concentration ratio must be between 1:10 and 10:1

First if 10 acid : 1 base:

pKa = pH - log { a(A-) / a(HA) }

= pH - log { 1 / 10 }

= pH - (-1) = pH + 1

Second if 1 acid : 10 base:

pKa = pH - log { a(A-) / a(HA) }

= pH - log { 10 / 1 }

= pH - 1

The buffer pH range for any weak acid with its conjugate is pKa 1

The buffer pH range for any weak base with its conjugate is pKa(conjugate) 1, (pOH range: pKb 1).

Making up a Buffer Solution.

1. From the given pH for the buffer solution choose an acid - base conjugate pair to fit the relationship:

pH (of buffer) = pKa 1

or pOH(of buffer) = pKb 1

The best value is when pH = pKa or pOH = pKb.

2. Calculate the actual ratio of concentrations needed for the buffer solution using

pKa = pH - log { a(A-) / a(HA) }

or

pKb = pOH - log { a(HB+) / a(B) }

Note that these concentrations are strictly those at equilibrium, not the initial concentrations. However, we have noted in calculations with a small K value that in {[initial] - x} the x can be ignored. This is happening here, so the concentration terms used in these equations are the initial concentrations.

3. Choose a ratio which gives both concentrations between 0.100 M and 1.00 M.

Example 1: Make up 1 L of a Buffer Solution to pH = 5.5

1. Choose an appropriate acid.

For pH = 5.5, the pKa of the acid must lie between 4.5 and 6.5 and preferably be close to 5.5.

Propanoic acid has pKa = 4.89.

2. Calculate the ratio of the concentrations of the acid and its conjugate:

pKa = pH - log { a(A-) / a(HA) }

4.89 = 5.5 - log { a(C2H5COO-) / a(C2H5COOH)}

so: log { a(C2H5COO-) / a(C2H5COOH)} = 0.61

and: a(C2H5COO-) / a(C2H5COOH) = 4.07

3. Choose an actual concentration ratio for the acid and its salt:

[C2H5COO-] = 0.407 M

[C2H5COOH] = 0.100 M

Answer: make up a solution which is 0.407 M in sodium propanoate and 0.100 M in propanoic acid.

Example 2: show how to make a buffer solution with pH = 8.3.

1. Since this is a basic buffer, use a base and its conjugate.

pOH(of buffer) = pKb   1

5.7 = pKb   1

Hydrazine has pKb = 5.89

2. Calculate the relative concentrations:

pKb = pOH - log { a(HB+) / a(B) }

5.89 = 5.7 - log { a(HB+) / a(B) }

-0.19 = log { a(HB+) / a(B) }

a(HB+) / a(B) = 0.65

3. Choose actual concentrations to fit this ratio:

[NH2NH3+ Cl-] = 0.325 M

[NH2NH2] = 0.500 M

Make up a solution which is 0.325 M in [NH2NH3+ Cl-] and 0.500 M in [NH2NH2].

Making a Buffer by Partial Titration. During a titration, an acid (or a base) is being converted into its conjugate. Therefore a buffer solution exists for a good part of this titration:

To create the buffer at pH = 5.5 using propanoic acid and sodium hydroxide, the titration reaction has to go until the ratio of propanoate to propanoic acid is 4.07 (previously calculated).

Sarting with 1.00 L of 1.00 M propanoic acid:

 C2H5COOH + NaOH CH3COO- Na+ + H20 I 1.00 mol 0 0 C (-1)x (x added) (+1)x End (1.00 - x) mol 0 x mol

But x / (1 - x) = 4.07 (again, volumes cancel, so change in volume is irrelevant)

so x = 0.803

Therefore, adding 803 mL of NaOH to the 1.00 L of 1.00 M propanoic acid will give a buffer with the required ratio of concentrations.