# Tough question from the 2014 CHEM 13 NEWS Exam

This article focuses on question #10 from the 2014 CHEM 13 News Exam, which is reproduced below. The number below each response is the percentage of students selecting that response. The correct answer is E.

What is the pH of pure water at 37 oC? At 37 oC, Kw = 2.4×10−14

1. 6.90 (4.5%)
2. 7.10 (5.4%)
3. 7.19 (5.5 %)
4. 7.00 (20.1%)
5. 6.81 (32.8%)

With a point biserial coefficient of 0.566, this question was, from a statistical perspective, the one that best discriminated between the “stronger” and “weaker” students. (The point biserial coefficient for a given question can have a value between −1 and +1 and measures the correlation between how students perform on that question and on the entire exam; the more positive the value, the stronger the correlation.) Only one-third of students answered this question correctly, with a significant fraction (31.8%) not entering any answer. Twenty percent of students chose answer D, which is the pH of pure water at 25 oC.

The quantity Kw is the ion product for water and the (thermodynamic) equilibrium constant for the reaction describing the self-ionization of water:

2 H2O(l) ⇌ H3O+(aq)  +  OH(aq),

or H2O(l) ⇌ H+(aq)  +  OH(aq).

If we choose to represent the self-ionization reaction using the first of these two equations, we can write

Kw = [H3O+]eq [OH ]eq.

From this, we have

[H3O+ ] = Kw1/2 = 1.5×10−7 mol L−1 and
pH = −log10[H3O+ ] = 6.81.

Another approach is to set up the following equilibrium summary (an “ICE table”).

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

 Initial: excess 0 mol L−1 0 mol L−1 Change: (− x) + x mol L−1 + x mol L−1 Equilibrium: excess x mol L−1 x mol L−1

In pure water, we must have [H3O+ ] = [OH]  and so,

Kw = [H3O+]eq2.

The ICE table indicates that [H3O+ ]eq = [OH]eq  = x mol L−1. The value of x is calculated by solving the equation Kw = x2, using the given value of Kw.

The ion product for water and the self-ionization of water can be used by teachers to emphasize the following acid-base concepts.

Differentiating between acidic, basic and neutral solutions:  The classification of a solution as acidic, basic or neutral is made either by comparing the relative amounts of H+ and OH, or by comparing the pH of a solution to that of pure water at the same temperature. Table 1 summarizes various criteria that may be used to decide whether a solution is acidic, basic or neutral.

### Table 1

 Acidic solution [H3O+ ] > [OH− ] pH < ½ pKw Basic solution [H3O+ ] < [OH− ] pH > ½ pKw Neutral solution [H3O+ ] = [OH− ] pH = ½ pKw

To use the criteria involving pH, we must first evaluate ½ pKw, where pKw = −log10 Kw. The quantity ½ pKw is the pH of pure water. The value of Kw varies between 1.14×10−15 at 0 oC and 5.45×10−13 at 100 oC, and so the pH of pure water decreases from 7.47 to 6.13 as the temperature increases from 0 oC to 100 oC. Although the pH of water decreases as the temperature increases, it is incorrect to say that water is more acidic at higher temperatures than at lower temperatures. The following question, a variation of the question above, could be used to assess whether students understand these concepts. The answer to the question below is that X is a base.

An aqueous solution of a substance X has a pH of 7.00 at 37 oC. Given that Kw = 2.4×10−14 at 37 oC, decide whether X is an acid, a base, or neither an acid nor a base.

Calculating Ka and Kb values of water:  In the reaction describing the self-ionization of water, one water molecule acts as an acid and the other acts as a base. In other words, the H2O molecule is amphiprotic.

A water molecule is amphiprotic because it contains a hydrogen atom bonded to an electronegative atom, as well as an electronegative atom with a lone pair of electrons that can be used to bind a proton. Given that H2O is amphiprotic, we should be able to quantify the acid and base strength of H2O by calculating values of the ionization constants, Ka and Kb.

By definition, Ka for an acid “HA” is equal to

[H3O+]eq[A ]eq/[HA] eq and

Kb for a base “B” is equal to [BH+]eq[OH ]eq/[B]eq.

By replacing HA and B with H2O in these expressions, we obtain

Ka(H2O) = Kb(H2O) = [H3O+ ]eq[OH ]eq/[H2O]eq.

We have [H3O+ ]eq[OH ]eq = Kw and

[H2O]eq = ρ(H2O)/M(H2O), where ρ(H2O) is the density of water (in grams per litre) and

M(H2O) = 18.016 g mol−1.

Therefore, Ka(H2O) = Kb(H2O) = KwM(H2O)/ρ(H2O).

At 25 oC, Kw = 1.0×10−14 and ρ(H2O) = 1.0×103 g L−1, so

Ka(H2O) = Kb(H2O) = 1.8×10−16.

The corresponding pKa and pKb values, defined as

pKa = −log10Ka and pKb = −log10Kb,

are both equal to 15.74.