Tough questions series – 2014 CHEM 13 NEWS Exam

(This is a reprint from the February 2015 issue of Chem 13 News, pages 14-15.)

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, K_w = 2.4×10⁻¹⁴ 

A   6.90 
     (4.5%)

B   7.10
     (5.4%)

C   7.19
     (5.5 %)

D   7.00
     (20.1%)

E   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 K_w 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 

K_w = [H₃O⁺]_eq [OH⁻]_eq. 

From this, we have

[H₃O⁺] = K_w^1/2 = 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”).

In pure water, we must have [H₃O⁺] = [OH⁻]  and so, K_w = [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 K_w = x², using the given value of K_w.

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 pK_w  

Basic solution         [H₃O⁺] < [OH⁻]   pH > ½ pK_w

Neutral solution      [H₃O⁺] = [OH⁻]   pH = ½ pK_w

To use the criteria involving pH, we must first evaluate ½ pK_w, where pK_w = −log¹⁰K_w. The quantity ½ pK_w is the pH of pure water. The value of K_w 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 K_w = 2.4×10⁻¹⁴ at 37 ^oC, decide whether X is an acid, a base, or neither an acid nor a base.

Calculating K_a and K_b 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, K_a and K_b.

By definition, K_a for an acid “HA” is equal to [H₃O⁺]_eq[A⁻]_eq/[HA]_eq and 

K_b 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 K_a(H₂O) = K_b(H₂O) = [H₃O⁺]_eq[OH⁻]_eq/[H₂O]_eq. 

We have [H₃O⁺]_eq[OH⁻]_eq = K_w 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, K_a(H₂O) = K_b(H₂O) = K_wM(H₂O)/ρ(H₂O). 

At 25 ^oC, K_w = 1.0×10⁻¹⁴ and ρ(H₂O) = 1.0×10³ g L⁻¹, so 
K_a(H₂O) = K_b(H₂O) = 1.8×10⁻¹⁶. 

The corresponding pK_a and pK_b values, defined as 

pK_a = −log₁₀K_a and pK_b = −log₁₀K_b, are both equal to 15.74. 

A follow up question is provided below. The answers are: 

      K_a = 55.5 and pK_a = −1.74 for H₃O⁺ 
and K_b = 55.5 and pK_b = −1.74 for OH⁻. 

For more information, see A.M. de Lange and J.H. Potgieter, Journal of Chemical Education, 68, 304, 1991.

Given that K_w = 1.0×10⁻¹⁴ and p(H₂O) = 1.0×10³ g/L at 25 ^oC, calculate K_a and pK_a for H₃O⁺ and K_b and pK_b for OH⁻.

The K_a values for H₂O and H₃O⁺, for example, can be used to predict what happens when a compound is added to water. If K_a for the compound is less than that of water, then the compound will not ionize and the solution will be pH neutral. If K_a is greater than that of water, but less that of H₃O⁺, then the compound will behave as a weak acid. If K_a is greater than that of H₃O⁺, then the compound will behave as a strong acid.