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Acids, Bases And pH

Contents

• Ionisation of water
• pH
• Acids and bases
• pKa and ionisation
• Henderson-Hasselbalch and buffers

Ionisation of water

Water is capable of reacting with itself to produce hydroxide ions (OH⁻) and 'protons' (H⁺). These protons are actually actually oxonium ions (H₃O⁺), but the H⁺ thing is so ingrained in the literature, it's rather hard to change now. We will use them interchangeably.

H₂O + H₂O ⇌ H₃O⁺ + OH⁻

H₂O ⇌ H⁺ + OH⁻

The ion product of water, K_w is derived from the equilibrium constant K_eq for this reaction:

K_eq = [H₃O⁺] [OH⁻] ⁄ [H₂O]² = 3.23 × 10⁻¹⁸

The acid dissociation constant is therefore:

K_a = [H₃O⁺] [OH⁻] ⁄ [H₂O] = 1.8 × 10⁻¹⁶ M

The value of 1.8 × 10⁻¹⁶ is found experimentally. The concentration of water in a solution, [H₂O], is more-or-less constant at 55.56 M at STP: the density of water is 1000 g L⁻¹ and the RMM of water is 18 g mol⁻¹, hence the concentration of water is:

1000 g L⁻¹ ⁄ 18 g mol⁻¹ = 55.56 mol L⁻¹ = 55.56 M.

Because of this constancy, we can define the ion product of water, K_w as [H₂O] × K_a:

K_w = [H₃O⁺] [OH⁻] = 1.0 × 10⁻¹⁴ M²

So the ion product of water is 10⁻¹⁴ M². In pure water, the number of 'protons' and hydroxide ions must be exactly equal (since they can only be produced in a 1:1 ratio from pure water), hence:

[H₃O⁺] = [OH⁻] = √10⁻¹⁴ M² = 10⁻⁷ M.

So, protons are present at only 0.1 µM, but they are very important as they determine the acidity or basicity of a solution. The ion product of water is a constant, regardless of how much acid, alkali, or what-have-you we add to the water. This is very important: no matter what we add to the water (within limits), the product of the concentrations of protons and hydroxide ions remains the same. This has important implications on the pH of water solutions of acids and bases, as we will see:

pH

The pH is defined as minus log to the base ten of the molar hydrogen ion concentration.

pH = − log₁₀ [H⁺]

pH is therefore a logarithmic scale of hydrogen ion concentration, which runs from 0 to 14 in practice:

• 0 is extremely acidic (e.g. 1 M hydrochloric acid).
• Stomach pH 1.7
• Orange juice pH 2.5
• 7 is neutral.
• Blood pH 7.4
• Oven cleaner pH 13.5
• 14 is extremely basic (e.g. 1 M sodium hydroxide).

Calculating pH is easy if we know the hydrogen ion concentration. In pure water, we just showed that [H⁺] = 10⁻⁷ M. Hence:

pH = − log₁₀ [H⁺] = − log₁₀ 10⁻⁷ = 7

Which is obvious really. In 1 M HCl, which being a very strong acid, is nearly all ionised to H⁺ and Cl⁻ ions, the proton concentration is 1 M, hence the pH is:

pH = − log₁₀ [H⁺] = − log₁₀ 1 = 0

To determine the pH of a 1 M solution of NaOH, which is a strong base and almost entirely ionised to 1 M OH⁻ ions, we need to remember that the ion product is a constant, i.e. [H⁺][OH⁻] = 10⁻¹⁴. Hence, if we know [OH⁻] is 1 M, then the proton concentration and pH are:

[H⁺] = 10⁻¹⁴ ⁄ [OH⁻] = 10⁻¹⁴ ⁄ 1 = 10⁻¹⁴

pH = − log₁₀ [H⁺] = − log₁₀ 10⁻¹⁴ = 14

The basicity of ammonia, NH₃, is due to the fact that it removes protons from solution to form ammonium ions NH₄⁺. Hence to maintain the ion product constant, the solution becomes highly enriched with hydroxide ions.

Acids and bases

There are several definitions of what acids and bases are. The oldest was that acids are those things that dissociate to produce protons, and bases are things that dissociate to form hydroxide ions. This makes ammonia something of a problem. Hence, a more modern treatment (Brønsted-Lowry theory) defines acids as proton donors, and bases as proton acceptors. In the equation:

HA ⇌ H⁺ + A⁻

HA is defined as an acid, and A⁻ is its conjugate base. In the equation:

B⁻ + H⁺ ⇌ BH.

B⁻ is defined as a base, and BH is its conjugate acid. Whenever we have an acid/base reaction, there is always a base and an acid acting as a conjugate pair. Another definition of acids and bases (Lewis theory) states that acids are lone pair acceptors, and bases are lone pair donors (e.g. OH⁻ is a base as it has a three lone pairs to donate to protons, and H⁺ is an acid because it can accept a lone pair from NH₃ or OH⁻ to form ammonium or water). We won't use this idea here, but it's worth knowing as it unifies acid-base reactions with complexation.

The pK_a of an acid is defined as the negative of the base ten logarithm of the equilibrium constant for the reaction, which itself is called K_a:

HA ⇌ H⁺ + A⁻

K_a = [H⁺] [A⁻] ⁄ [HA].

pK_a = − log₁₀ K_a

Strong acids have a low pK_a, and weak acids have a high pK_a. There is also a pK_b (and a pOH) which are to bases what pK_a and pH are to acids:

B + H⁺ ⇌ BH⁺

K_b = [BH⁺] ⁄ [B] [H⁺].

pK_b = − log₁₀ K_b

pOH = − log₁₀ [OH⁻]

Generally, in biochemistry, it is easier to consider just the acid versions. For pOH and pH, there is a simple relationship from the ion product:

pK_w= − log₁₀ K_w= 14

pK_w= − log₁₀ [H⁺] [OH⁻] = − log₁₀ [H⁺] − log₁₀ [OH⁻] = pH + pOH = 14

So if you ever need to know the pOH of a solution, it's just 14 minus the pH (and vice versa). Also, rather than talking about the K_b of the reaction:

B + H⁺ ⇌ BH⁺

K_b = [BH⁺] ⁄ [B] [H⁺].

It's usually easier to just consider the K_a of the reverse:

BH⁺ ⇌ H⁺ + B

K_a = [H⁺] [B] ⁄ [BH⁺].

It's fairly obvious that the K_a of the reverse reaction is the reciprocal of the K_b of the original reaction, and therefore pK_a is minus pK_b:

K_a = 1 ⁄ K_b and K_b = 1 ⁄ K_a

pK_a = − log₁₀ K_a = − log₁₀ (1 ⁄ K_b) = log₁₀ K_b = − pK_b

Some amino acids are conventionally termed 'basic', that is they tend to accept protons at physiological pH. Hence, they are mostly charged (RNH₃⁺). The pK_a for these amino acids would therefore refer to the dissociation of the proton from the charged form, i.e. RNH₃⁺ ⇌ RNH₂ + H⁺.

Water itself is a special case of a compound that can act both as an acid and a base. Here are the two acid/base half-reactions for water:

H₂O (conjugate acid) ⇌ H⁺ + OH⁻ (conjugate base)

H₂O (conjugate base) + H⁺ ⇌ H₃O⁺ (conjugate acid)

These two half reactions add up to the ionisation of water:

H₂O + H₂O ⇌ H₃O⁺ + OH⁻

Water can act as both an acid and a base, and so can any other compound! It all depends on the relative strength of the acid and base in the reaction, i.e. on the pK_a of the two reactants:

HCl (strong acid) + HCOOH (weak acid) ⇌ HC(OH)₂⁺ + Cl⁻

HCOOH (weak acid) + H₂O (very weak acid) ⇌ HCOO⁻ + H₃O⁺

pK_a and ionisation

The pK_a can be thought of as a personal scale of acidity for a compound. If you add an acid to a pH buffer that maintains the pH at a certain value (we'll talk about how to do this presently), you can compare the pK_a of the acid and the pH of the buffer to discover how ionised the acid is.

If the buffer's pH is lower than the acid's pK_a, the buffer ' looks acidic' to the acid, and this will stop the acid from ionising (there are already plenty of protons about, and the acid doesn't bother adding to them). If the pH is higher than the acid's pK_a, then the solution 'looks alkaline' (i.e. there are lots of hydroxide ions about), and these hydroxide ions will strip protons off the acid, ionising it. So, consider formic acid (HCOOH), which has a pK_a of 3.5. If you add it to buffer at pH 3.5, half the formic acid will be in the form HCOOH, the conjugate acid, and the other half will be in the ionised form HCOO⁻ , the conjugate base. We can use the Henderson-Hasselbalch equation (soon) to determine the percentage ionisation:

• At pH 1.5, it's 1% COO⁻.
• At pH 2.5, it's 10% COO⁻.
• At pH 3.5, it's 50% COO⁻.
• At pH 4.5, it's 90% COO⁻.
• At pH 5.5, it's 99% COO⁻.

So if you expose formic acid to a buffer containing an even stronger acid (e.g. sulfuric acid), with a very low pK_a and very low pH, the formic acid/formate conjugate pair will act as a proton acceptor, i.e. as a base. But add it to a solution buffered to a high pH, and it will donate protons, i.e. act as an acid. Formic acid's acidity is purely relative.

Why does pK_a matter for biochemists? Because it matters for proteins. Enzymes often work by ionic interactions with substrates. e.g. an amino acid in the active site must be positively charged to bind a negatively charged substrate. This is why enzymes have an optimum pH. Whether or not an amino acid is protonated, and hence, what the charge on the amino acid is, depends on the pH of the surrounding solution.

Henderson-Hasselbalch and buffers

The Henderson-Hasselbalch equation is extremely useful for calculating the expected pH of a solution of an acid and its conjugate base. The equation is:

pH = pK_a + log₁₀ [A⁻] ⁄ [HA].

The derivation of the equation is quite simple:

K_a = [H⁺] [A⁻] ⁄ [HA].

Rearrange this.

[H⁺] = K_a [HA] ⁄ [A⁻] .

Take − log₁₀ of both sides and simplify (note the flip of the HA and A⁻, this is due to the nature of logarithms).

− log₁₀ [H⁺] = − log₁₀ ( K_a [HA] ⁄ [A⁻] ).

− log₁₀ [H⁺] = − log₁₀ K_a + log₁₀ ( [A⁻] ⁄ [HA] ).

Two of these numbers we've met before, they are pH and pK_a.

pH = pK_a + log₁₀ ( [A⁻] ⁄ [HA] ).

This is the Henderson-Hasselbalch equation. Let's use it to make a buffer. A buffer is a solution that resists changes in its pH even when you add quite large volumes of a strong acid or base to it. They are obviously useful in biochemistry for maintaining controlled conditions during experiments. The human body is also pH buffered, so you don't dissolve when you drink orange juice. There are two ways to make a buffer, both of which are equivalent. We can:

• Partially neutralise a weak acid/base with a strong base/acid.
• Mix the salt of a weak acid/base with the weak acid/base.

These are equivalent because in both cases we end up with a solution of a weak acid/base and its salt. In the latter case, we make it directly by mixing the acid/base and salt, and in the former, we create the salt in situ by neutralising some of the acid/base. The former case is termed titration. Titrations display the character of the buffer we wish to make: the rate of change of pH is slowest at the pK_a of the weak acid we have made the buffer from.

pK_a = pH when [HA] = [A⁻], i.e. 50% neutralised.

The solution is well buffered at this pH, but buffer action is only effective +/− 1 pH unit around the pK_a of the acid used.

An example calculation. I want to make 1 L of 0.1 M phosphate buffer from K₂HPO₄ and KH₂PO₄ at pH 7.2 . How much of each salt do I require? The pK_a of phosphate (2^nd ionisation) is 6.8.

First plug the figures into Henderson-Hasselbalch. Note that HA is the acid (the phosphate salt with more hydrogen, i.e. KH₂PO₄), and A⁻ is the base (the phosphate salt with less hydrogen, i.e. K₂HPO₄).

7.2 = 6.8 + log₁₀ ( [A⁻] ⁄ [HA] ).

Rearrange this and use your calculator:

10 ^{(7.2 − 6.8)} = [A⁻] ⁄ [HA] = 2.5

We also know the concentration of base plus acid must be 0.1 M (it's a 0.1 M buffer after all). So:

[A⁻] + [HA] = 0.1 M

[A⁻] = 0.1 − [HA]

If we substitute this definition of [A⁻] into the previous equation:

( 0.1 − [HA] ) ⁄ [HA] = 2.5

Rearrange this:

0.1 − [HA] = 2.5 [HA]

0.1 = 2.5 [HA] + [HA]

0.1 = (2.5 + 1) [HA].

[HA] = 0.1 ⁄ (2.5 + 1) = 0.028 M.

And since they add to 0.1, [A⁻] must be 0.072 M. To get the actual amounts we need, we need to use mole to gram conversions.

KH₂PO₄ has RMM of 136 g mol⁻¹

K₂HPO₄ has RMM of 174 g mol⁻¹

KH₂PO₄ : 136 g mol⁻¹ × 0.028 mol L⁻¹ × 1 L = 3.8 g

K₂HPO₄ : 174 g mol⁻¹ × 0.072 mol L⁻¹ × 1 L = 12.5 g

Calculation complete. If you don't remember anything else, try to remember the Henderson-Hasselbalch equation, and if your maths is iffy, try to remember the rearranged equation:

[HA] = [buffer] ⁄ ( 1 + 10^{(pH − pKa )} ).

Where [buffer] is the overall buffer molarity. Once you know [HA], you can work out how many grams or moles of the salts you need, or work out what volume of standard solutions of salts you need to mix.

Test yourself

1. What does the ion product of water represent?
2. Write an equation to relate the pH and pOH of a solution.
3. What is the pH of a 10 mM solution of nitric acid?
4. Write an equation to show the reaction between ethanoic acid and sodium hydroxide, labelling the conjugate pairs.
5. What is the pH of a buffer made from serine, whose protonated amino group has a pKa of 9.2, when the acid form is present at 20 mM and the base at 15 mM?

Answers