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 K_b 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:

K_a(HA) * K_b(A-) = K_w = 1 x 10^-14

or (taking logs)

pK_a + pK_b = pK_w = 14 for conjugates.

1. For conjugate bases of HA.

pK_a(HA) + pK_b(A^-) = 14.

or

pK_b(A^-) = 14 - pK_a(HA)

Rules:

If pK_a or pK_b > 14, the acid or base will not react significantly with water.

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

The consequence of these:

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

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

2. For conjugate acids of B:

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

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

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:

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.

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 Brønsted acids:

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

We will ignore this possibility.

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

pK_a (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

K_b = a(CH₃COOH) * a(OH^-) / a(CH₃COO^-)

K_b = x * x / (0.100 - x)

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

pK_b(CH₃COO^-) = 14 - pK_a(CH₃COOH)

= 9.25

and

K_b(CH₃COO^-) = 5.62 x 10^-10.

Substituting:

K_b = 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.

pK_b 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.

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.

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.

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 pK_a at least 2 units higher than the pK_a 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 pK_b at least 2 units higher than the pK_b 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.

For strong acid - strong base titrations:

any indicator will do (all are weak acids).

For weak acid - strong base titrations:

use indicator with a pK_a = 8 or more

for example: phenolphthalein (pK_a = 9.4)

For strong acid - weak base titrations:

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

for example: methyl orange (pK_a = 3.4,

pK_b of the conjugate ion = 10.6)

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

K_a = a(H₃O⁺) * a(In^-) / a(HIn)

in logarithms:

pK_a = 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:

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

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

= pH - (-1)

= pH + 1

colour 2 pK_a = 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 pK_a - 1 to pK_a + 1

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

pK_a + 1 to pK_a - 1.

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

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 K_a for an acid:

K_a = a(H₃O⁺) * a(A^-) / a(HA)

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

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

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

(pK_a(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

K_a = {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

K_b = a(CH₃COOH) * a(OH^-) / a(CH₃COO^-)

K_b = 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 = pK_a + 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 = pK_a + log { a(CH₃COO^-) / a(CH₃COOH) }

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

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

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

pK_a = 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 K_a for nitrous acid.

Answer:

	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

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

pK_a = 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 pK_a(HNO₂) = 3.14.

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

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

Take the situation exactly half way through the titration. At this point x = ½ a, so:

pK_a = pH - log { ½ a / (a - ½ a)}.

pK_a = pH - log { 1 }.

pK_a = pH

Half way through a titration, the pK_a 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.

From the definition of Ka for an acid:

K_a = a(H₃O⁺) * a(A^-) / a(HA)

or in logs: pK_a = 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:

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

= pH - log { 1 / 10 }

= pH - (-1) = pH + 1

Second if 1 acid : 10 base:

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

= pH - log { 10 / 1 }

= pH - 1

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

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

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

pH (of buffer) = pK_a 1

or pOH(of buffer) = pK_b   1

The best value is when pH = pK_a or pOH = pK_b.

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

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

or

pK_b = 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.

1. Choose an appropriate acid.

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

Propanoic acid has pK_a = 4.89.

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

pK_a = 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.

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

pOH(of buffer) = pK_b   1

5.7 = pK_b   1

Hydrazine has pK_b = 5.89

2. Calculate the relative concentrations:

pK_b = 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

Answer:

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

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.