Chemistry 112

Strong Acids and Bases in Water


In aqueous solution, the consequences of any acid–base chemistry are observed by the changes in the properties of water.

The autoionization of water, in the Brønsted–Lowry context is given by:

H2O(l) + H2O(l) H3O+(aq) + OH(aq)

Water is a weaker base than hydroxide and water is a weaker acid than hydronium ion. (Why?)

Thus, the equilibrium predominates towards the reactants side.

The mass action expression for this equilibrium is:

Kc = [H3O+]e[OH]e = Kw

Kw is the equilibrium constant for the autoionization of water.

At 25 oC, Kw = 1.0×10–14

Memorize this forever!



Examples

What are the concentrations of hydroxide and hydronium ion in pure water at 25 oC?



Now suppose we prepare a 0.010 M solution of nitric acid in water. What are the hydronium ion and hydroxide ion concentrations at 25 oC?

There are now two reactions to consider:

HNO3(aq) + H2O(l) H3O+(aq) + NO3 (aq)

2 H2O(l) H3O+(aq) + OH (aq)

The strong acid contributes 0.010 M hydronium ion because the reaction goes to completion.

This causes a shift in the water ionization equilibrium towards reactants so the amount of H3O+ and OH contributed by the water is less than 1.0×10–7 M. (LeChatelier's Principle)

Since the amount of hydronium ion contributed by the water is so small, we can ignore it:

[H3O+]e = 0.010 + (<1.0×10–7) = 0.010 M

The hydroxide ion concentration is found from Kw:

Kw = [H3O+]e[OH]e = 1.0×10–14

[0.010][OH]e = 1.0×10–14

[OH]e = 1.0×10–12 M

The effect of the strong acid is to increase the concentration of hydronium ion and suppress the concentration of hydroxide ion.



Suppose a 1.0×10–6 M solution of NaOH is prepared. What are the hydronium ion and hydroxide ion concentrations at 25 oC?

Again, we need to consider two reactions:

NaOH(aq) Na+(aq) + OH(aq)

2 H2O(l) H3O+(aq) + OH(aq)

The strong base contributes 1.0×10–6 M hydroxide.

The water contributes less than 1.0×10–7 M hydroxide, so the total is

[OH] = 1.0×10–6 + <1.0×10–7 ~ 1.0×10–6 M

Kw = [H3O+]e[OH]e = 1.0×10–14

[H3O+]e[1.0×10–6] = 1.0×10–14

[H3O+]e = 1.0×10–8 M



Suppose that a 1.0×10–7 M solution of KOH were prepared. What are the hydronium ion and hydroxide ion concentrations at 25 oC?

In general, if the concentration of a strong acid or strong base exceeds 1.0×10–6 M, then the effects of the water autoionization equilibrium can be ignored.

If the concentration of the strong acid or strong base is less than 1.0×10–6 M, then a full equilibrium calculation must be done.



pH and pOH

The range of hydronium ion or hydroxide ion concentrations is large, typically from 10–1 to 10–14 M. All these exponents are annoying, so we have defined a scale that measures hydronium ion concentration in terms of a smaller range of numbers. This is the pH scale:

pH = –log[H3O+]

In pure water:

[H3O+] = 1.0×10–7 so the pH is

pH = –log[H3O+] = –log(1.0×10–7) = 7.00

(A note on significant figures when using logs.)

Values of the pH below 7 indicate an excess of hydronium ion over hydroxide ion. This is referred to as an acidic solution.

Values of the pH greater than 7 indicate an excess of hydroxide ion over hydronium ion. This is referred to as a basic solution.



Examples

Find the pH of a 0.0050 M HBr solution at 25 oC.

HBr is a strong acid, so:

HBr(aq) + H2O(l) H3O+(aq) + Br(aq)

Since the concentration of the strong acid exceeds 1.0×10–6 M, we can ignore the contribution from the water autoionization equilibrium. Thus,

[H3O+]e = 5.0×10–3M

pH = –log[H3O+] = –log(5.0×10–3) = 2.30


7.32×10–5 moles of NaOH is added to 2.0 L of water. What is the pH at 25 oC?

This is a strong base, so:

NaOH(aq) Na+(aq) + OH (aq)

[OH] = 7.32×10–5mol/2.0 L = 3.7×10–5 M

Since the concentration of the strong base exceeds 1.0×10–6 M, we can ignore the contribution from the water autoionization equilibrium. Thus,

[H3O+]e[OH]e = 1.0×10–14

[H3O+]e[3.7×10–5] = 1.0×10–14

[H3O+]e = 2.7×10–10 M

pH = –log[H3O+] = –log(2.7×10–10) = 9.57


Equivalently, we can define pOH:

pOH = –log[OH]

At 25 oC, pH and pOH are related by

pH + pOH = 14.00 (only at 25 oC)

(The derivation for this expression is shown here.)

Thus, in the NaOH example:

pOH = –log[OH] = –log(3.7×10–5) = 4.43

Note, 4.43 + 9.57 = 14.00, as expected.