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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听^oC? At 37听^o颁,听K_w听= 2.4脳10^鈭�14

1. 6.90听(4.5%)
2. 7.10 (5.4%)
3. 7.19听(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 鈥渟tronger鈥� and 鈥渨eaker鈥� students. (The point biserial coefficient for a given question can have a value between 鈭�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听^oC.

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

2 H₂O(l) 鈬� H₃O⁺(aq)听 +听 OH^鈭�(aq),

or H₂O(l) 鈬� H⁺(aq)听 +听 OH^鈭�(aq).

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

K_w听= [H₃O⁺]_eq听摆翱贬^鈭�听闭_eq.

From this, we have

[H₃O⁺听闭 =听K_w^1/2听= 1.5脳10^鈭�7听mol L^鈭�1听补苍诲
pH = 鈭抣og₁₀[H₃O⁺听闭 = 6.81.

Another approach is to set up the following equilibrium summary (an 鈥淚CE table鈥�).

2 H₂O(l) 听 听 听 听 听 鈬� 听 听 听 听 听 听 听 听 H₃O⁺(aq) 听 听 听 听 听 听 听 听 听 听 听 听+ 听 听 听 听 听 听 听 听 听 听 听 听OH^鈭�(aq)

In pure water, we must have [H₃O⁺听闭 = [OH^鈭�] 听补苍诲 so,

K_w听= [H₃O⁺]_eq².

The ICE table indicates that [H₃O⁺听闭_eq听= [OH^鈭�]_eq听听= x mol L^鈭�1. The value of x is calculated by solving the equation听K_w听= x², using the given value of听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: 听The classification of a solution as acidic, basic or neutral is made either by comparing the relative amounts of H⁺听补苍诲 OH^鈭�, 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

To use the criteria involving pH, we must first evaluate 陆 pKw, where pK_w听= 鈭抣og₁₀听K_w. The quantity 陆 pK_w听is the pH of pure water. The value of听K_w听varies between 1.14脳10^鈭�15听at 0听^oC and 5.45脳10^鈭�13听at 100听^oC, and so the pH of pure water decreases from 7.47 to 6.13 as the temperature increases from 0听^oC to 100听^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听^oC. Given that听K_w听= 2.4脳10^鈭�14听at 37听^oC, decide whether X is an acid, a base, or neither an acid nor a base.

Calculating听K_a听补苍诲听K_b听values of water:听 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,听K_a听补苍诲听K_b.

By definition,听K_a听for an acid 鈥淗A鈥� is equal to

[H₃O⁺]_eq[A^鈭�听闭_eq/[HA]听_eq听补苍诲

K_b听for a base 鈥淏鈥� is equal to [BH⁺]_eq[OH^鈭�听闭_eq/[B]_eq.

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

K_a(H₂O) =听K_b(H₂O) = [H₃O⁺听闭_eq[OH^鈭�听闭_eq/[H₂O]_eq.

We have [H₃O⁺听闭_eq[OH^鈭�听闭_eq听=听K_w听补苍诲

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

M(H₂O) = 18.016 g mol^鈭�1.

Therefore,听K_a(H₂O) =听K_b(H₂O) =听K_wM(H₂O)/蚁(H₂O).

At 25听^o颁,听K_w听= 1.0脳10^鈭�14听补苍诲听蚁(H₂O) = 1.0脳10³听g L^鈭�1, so

K_a(H₂O) =听K_b(H₂O) = 1.8脳10^鈭�16.

The corresponding pK_a听补苍诲 pK_b听values, defined as

pK_a听= 鈭抣og₁₀K_a听补苍诲 pK_b听= 鈭抣og₁₀K_b,

are both equal to 15.74.

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

K_a听= 55.5 and pK_a听= 鈭�1.74 for H₃O⁺
and听K_b听= 55.5 and pK_b听= 鈭�1.74 for OH^鈭�.

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

Given that听K_w听= 1.0脳10^鈭�14听补苍诲听蚁(H₂O) = 1.0脳10³听g/L at 25听^oC, calculate听K_a听补苍诲 pK_a听for H₃O⁺听补苍诲听K_b听补苍诲 pK_b听for OH^鈭�.

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