Collections:Texts:Chemistry:14.2 The pH of Solutions of Weak Acids and Weak Bases

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14.2 THE pH OF SOLUTIONS OF WEAK ACIDS AND WEAK BASES

Weak Acids

When any weak acid, which we will denote by the general formula HA, is dissolved in water, the reaction

HA + H₂O H₃O⁺ + A^–      (14.4)

proceeds to only a limited extent, and we must allow for this in calculating the hydronium-ion concentration and hence the pH of such a solution. In general the pH of a solution of a weak acid depends on only two factors, the concentration of the acid, c_a, and the magnitude of an equilibrium constant K_a, called the acid constant, which measures the strength of the acid. The acid constant is defined by the relationship:

      (14.5)

(The acid constant is K_c multiplied by the constant concentration of water, as already defined in Sec. 13.2. It is also called the ionization constant or the dissociation constant of the acid.)

In Example 13.6 we showed how measurements of the conductivity of acetic acid solutions could be used to find the acid constant for acetic acid, K_a(CH₃COOH). You will recall from that example that K_a was only approximately a constant, varying from a value of 1.81 × 10^–5 mol dm^–3 in very dilute solutions to a value of 1.41 × 10^–7 mol dm^–3 in a 1M solution. A similar variation is found for other weak acids, so that most of the calculations we do using K_a are only approximate. Only two or possibly three significant figures should be retained. Table 14.2 gives the K_a values for a few selected acids arranged in order of their strength. It is at once apparent from this table that the larger the K_a value, the stronger the acid. The strongest acids, like HCl and H₂SO₄ have K_a values which are too large to measure, while another strong acid, HNO₃, has K_a value close to 20 mol dm^–3. Typical weak acids such as HF and CH₃COOH have acid constants with a value of 10^–4 or 10^–5 mol dm^–3. Acids like the ammonium ion, NH₄⁺, and hydrogen cyanide, HCN, for which K_a is less than 10^–9 mol dm^–3 are very weakly acidic.

Before we can go on to discuss how the hydronium-ion concentration and the pH of a solution of a weak acid depend on the concentration of the acid, we need to clarify a point of terminology. In order to do this let us take as an example a 0.0010 M solution of acetic acid. As we saw in Sec. 11.3, conductivity measurements show that only about 10 percent of the acid molecules have donated protons to water at any given time. We thus have a situation which can be summarized schematically in the following way:

      (14.6)

In such a solution there is some ambiguity as to what we mean by the phrase concentration of acetic acid. Do we mean 0.0010 mol dm^–3, or do we mean 90 percent of this value, namely, 0.0009 mol dm^–3? In order to resolve this difficulty, we will use the term stoichiometric concentration of acid and the symbol c_a to indicate the quantity 0.0010 mol dm^–3, that is, to indicate the total amount of acetic acid originally added per unit volume of solution. On the other hand we will use the term equilibrium concentration

TABLE 14.2 The Acid Constants for Some Acids at 25°C.

and the symbol [CH₃COOH] indicate the quantity 0.0009 mol dm^–3 that is, the final concentration of this species in the equilibrium mixture.

Let us now consider the general problem of finding [H₃O⁺]in a solution of a weak acid HA whose acid constant is K_a and whose stoichiometric concentration is c_a. According to the equation for the equilibrium,

HA + H₂O H₃O⁺ + A^–      (14.4)

for every mole of H₃O⁺ produced, there must also be a mole of A^– produced. At the same time a mole of HA and a mole of H₂O must be consumed. Since the volume which all these species occupy is the same, any increase in [H₃O⁺] must be accompanied by an equal increase in [A^–] and an equal decrease in [HA]. Consequently we can draw up the following table (in which equilibrium concentrations of all species have been expressed in terms of [H₃O⁺]:

*The hydronium-ion concentration actually increases from 10⁷ mol dm^-3 to the equilibrium concentration, and so the change in each of the concentrations is ± ([H₃O⁺ ] –10^-7 mol dm^-3). However, the concentration of hydronium ions produced by the weak acid is usually so much larger than –10^-7 mol dm^-3 that the latter quantity can be ignored. In the case of 0.0010 M acetic acid, for example, [H₃O⁺ ] ≈ 1 × 10^-4 mol dm^-3. Subtracting gives:

0.000 100 0 mol dm^-3

– 0.000 000 1 mol dm^-3

0.000 099 9 mol dm^-3

which is very close to 1 × 10^-4 mol dm^-3.

We can now substitute the equilibrium concentrations into the expression

      (14.7)

This could be solved for [H₃O⁺] by means of the quadratic formula, but in most cases a quicker approximate method is available. Since the acid is weak, only a small fraction of the HA molecules will have donated protons to form H₃O⁺ ions. Therefore [H₃O⁺] is only a small fraction of c_a and can be ignored when we calculate c_a – [H₃O⁺]. That is,

c_a – [H₃O⁺] ≈ c_a       (14.8)

Equation (14.7) then becomes

which rearranges to

[H₃O⁺]² ≈ K_ac_a

Taking the square root of both sides gives an important approximate formula:

[H₃O⁺] ≈       (14.9)

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EXAMPLE 14.5 Use Eq. (14.9) to calculate the pH of a 0.0200-M solution of acetic acid. Compare this with the pH obtained using the [H₃O⁺] of 5.92 × 10^–4 mol dm^–3 derived from accurate conductivity measurements.

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Solution From Table 14.2, K_a = 1.8 × 10^–5 mol dm^–3. Since c_a = 2.00 × 10^–2 mol dm^–3, we have

Using the accurate [H₃O⁺] from conductivity measurements,

pH = –log(5.92 × 10^–4) = –(0.772 – 4) = 3.228

Note that the approximate equation gives an [H₃O⁺] which differs by 1 in the second significant digit from the accurate value. The calculated pH differs by 1 in the second place to the right of the decimal―roughly the same as the accuracy of simple pH measurements.

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In a few cases, if the acid is quite strong or the solution very dilute, it turns out that Eq. (14.9) is too gross an approximation. A convenient rule of thumb for determining when this is the case is to take the ratio [H₃O⁺]/ c_a. If this is larger than 5 percent or so, we need to make a second approximation, and then the rules for successive approximations developed in the previous chapter can be applied. Equation (14.7) can be converted to a convenient form for successive approximations by multiplying both sides by c_a – [H₃O⁺]:

         [H₃O⁺]² = K_a(c_a - [H₃O⁺])

or      [H₃O⁺] = ...(14.10)

To get a second approximation for [H₃O⁺] we can feed the first approximation into the right side of this equation. The exact procedure is detailed in the following example.

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EXAMPLE 14.6 Find the pH of 0.0200 M HF using Table 14.2.

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Solution Using Eq. (14.9) we find

Checking we find

A second approximation is thus necessary. We feed our first approximation of [H₃O⁺]= 3.66 × 10^–2 mol dm^–3 into Eq. (14.10):

Taking a third approximation we find

Since this differs from the second approximation by less than 5 percent, we take it as the final result. The pH is

pH = –log(3.34 × 10^–3) = 2.48

Cross check: Since HF is a stronger acid than acetic acid, we expect this solution to have a lower pH than 0.02 MCH₃COOH and indeed it does.

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Weak Bases

The pH of a solution of a weak base can be calculated in a way which is very similar to that used for a weak acid. Instead of an acid constant K_a, a base constant K_b must be used. If a weak base B accepts protons from water according to the equation

B + H₂O BH⁺ + OH^–      (14.11)

then the base constant is defined by the expression

      (14.12)

A list of K_b values for selected bases arranged in order of strength is given in Table 14.3.

TABLE 14.3 The Base Constants for Some Bases at 25°C.

To find the pH we follow the same general procedure as in the case of a weak acid. If the stoichiometric concentration of the base is indicated by c_b, the result is entirely analogous to Eq. (14.7); namely,

      (14.13)

Under most circumstances we can make the approximation

c_b – [OH^–] ≈ c_b

in which case Eq. (14.13) reduces to the approximation

[OH^–] ≈       (14.14)

which is identical to the expression obtained in the acid case [approximation shown in Eq. (14.9)] except that OH^– replaces H₃O⁺ and b replaces a. Once we have found the hydroxide-ion concentration from this approximation, we can then easily find the pOH, and from it the pH.

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EXAMPLE 14.7 Using the value for K_b listed in Table 14.3, find the pH of 0.100 M NH₃.

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Solution It is not a bad idea to guess an approximate pH before embarking on the calculation. Since we have a dilute solution of a weak base, we expect the solution to be only mildly basic. A pH of 13 or 14 would be too basic, while a pH of 8 or 9 is too close to neutral. A pH of 10 or 11 seems reasonable. Using Eq. (14.14) we have

Checking the accuracy of the approximation, we find

The approximation is valid, and we thus proceed to find the pOH.

From which

pH = 14.00 – pOH = 14.00 – 2.87 = 11.13

This calculated value checks well with our initial guess.

Occasionally we will find that the approximation

c_b – [OH^–] ≈ c_b

is not valid, in which case we must use a series of successive approximations similar to that outlined above for acids. The appropriate formula can be derived from Eq. (14.13) and reads

[OH^–] ≈       (14.15)

Polyprotic Acids and Bases

In the case of polyprotic acids and bases we can write down an equilibrium constant for each proton lost or gained. These constants are subscripted 1, 2, etc., to distinguish them. For sulfurous acid, a diprotic acid, we can, for example, write

H₂SO₃ + H₂O H₃O⁺ + HSO₃^–

and

HSO₃^– + H₂O H₃O⁺ + SO₃^2–

The carbonate ion, CO₃^2–, is an example of a diprotic base for which the appropriate base constants are

CO₃^2– + H₂O HCO₃^– + OH^–

HCO₃^– + H₂O H₂CO₃ + OH^–

A general treatment of the pH of solutions of polyprotic species is beyond the scope of this text, but it is worth noting that in many cases we can treat polyprotic species as monoprotic. In the case of H₂SO₃, for example, K_a1 is very much larger than K_a2 indicating that H₂SO₃ is a very much stronger acid than HSO₃^–. This means that when H₂SO₃ is dissolved in water, we can treat it as a monoprotic acid and ignore the possible loss of a second proton. Solutions of salts containing the carbonate ion, such as N₂CO₃ or K₂CO₃ can be treated similarly.

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EXAMPLE 14.8 Find the pH of a 0.100-M solution of sodium carbonate, N₂CO₃. Use the base constant K_b1 = 2.10 × 10^–4 mol dm^–3_.

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Solution We ignore the acceptance of a second proton and treat the carbonate ion as a monoprotic base. We then have

Checking, we find that

so that our approximation is only just valid.

We now find

             pOH = –log(4.58 × 10^–3) = 2.34

while      pH = 14.00 – 2.34 = 11.66

Since the carbonate ion is a somewhat stronger base than NH₃, we expect a 0.1-M solution to be somewhat more basic, as actually found.

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A glance at Tables 14.2 and 14.3 reveals that most acid and base constants involve numbers having negative powers of 10. As in the case of [H₃O⁺] and [OH^–], then, it is convenient to define

...and...

Using these definitions, the larger K_a or K_b is (i.e., the stronger an acid or base, respectively), the smaller pK_aor pK_b will be. For a strong acid like HNO₃, K_a = 20 mol dm^–3 and

pK_a = –log 20 = – (1.30) = –1.3

Thus for very strong acids or bases pK values can even be negative.

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