(This is a reprint from the September 2009 issue of *Chem 13 News*, pages 10-11.)

Questions #12 and #24 from the 2009 CHEM 13 NEWS Exam challenged most students and quite a few teachers! Let’s focus first on question #12, which is reproduced below.

**What is the pH of 1.25×****10 ^{−7} mol L^{−1} HCl(aq)?**

The majority of students who answered this question recognized that HCl is a strong acid that dissociates completely in aqueous solution. Unfortunately, most of them calculated the pH incorrectly using pH = −log_{10}(1.25×10^{−7}) = 6.90. What is wrong with this approach? For all acid solutions, there are always two sources of H^{+} in solution: H^{+} from the acid and H^{+} from the self-ionization of water. The H^{+} that comes from the self-ionization of water is almost always neglected but it can't be neglected in very dilute solutions. For this question, there is only one equilibrium process to consider. The equilibrium summary is as follows.

*K*_{w} = 1.0×10^{−14} (at 298 K)

Reactants/Products | [H_{2}O(l)] |
[H^{+}(aq)] |
[OH^{−}(aq)] |
---|---|---|---|

Initial | - | C | 0 |

Change | - | + x | + x |

Equilibrium | - | C + x | x |

In the equilibrium summary, C is the initial concentration of HCl. Substituting [H^{+}] = C + x and [OH^{−}] = x into the expression [H^{+}][OH^{−}] = *K*_{w}, we obtain with some rearrangement the equation x^{2} + Cx − *K*_{w} = 0. The equation can be solved using the quadratic formula. Following this approach, we obtain the answer pH = 6.74. (Before doing this type of problem in the classroom, ask your students to calculate the pH of 1×10^{−8} mol/L HCl(aq). Most, if not all, will calculate an incorrect pH of 8.0. Obviously, that answer is not correct because it suggests that a basic solution can be made by dissolving a small amount of strong acid in water! Follow up with showing them how to calculate the correct answer, which is pH = 6.98. A very dilute solution of a very strong acid should be only very slightly acidic. I find that this is an interesting way to remind students about the self-ionization of water.)

Another approach is to use charge and mass balance equations, which I've touched on before. (See, for example, the May 2007 issue of *Chem 13 News*.) Even though there are various ionization processes occurring in solution (i.e., the ionization of HCl and the self-ionization of water), we should be able to keep track of mass and charge; mass is neither created nor destroyed and similarly, charge is neither created nor destroyed. When HCl is dissolved in water, it dissociates into H^{+} and Cl^{−} ions. All of the dissolved HCl ends up as an equivalent amount of Cl^{−}. Thus, the mass balance equation is C = [Cl^{−}]. (For HF(aq), the mass balance equation would be C = [HF] + [F^{−}] because HF is a weak acid and [HF] ≠ 0 at equilibrium. The right hand side of the mass balance equation for HF(aq) includes the concentrations of all fluorine containing species.) For HCl(aq), the charge balance equation is [H^{+}] = [Cl^{−}] + [OH^{−}]. Thus, the positive charge in solution must offset the negative charge. (Be careful when writing the charge balance equation for a solution containing ions with charges other than +1 or −1. See the May 2007 issue of *Chem 13 News* for more information on this point.) A close examination of the charge balance equation shows that [H^{+}] = [Cl^{−}] only if [OH^{−}] can be neglected. In this problem, [OH^{−}] can't be neglected. Using [OH^{−}] = *K*_{w}/[H^{+}] and [Cl^{−}] = C in the charge balance equation yields the equation [H^{+}]^{2} − C [H^{+}] – K_{w} = 0 which can be solved for [H^{+}] using the quadratic formula.

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Let’s now turn our attention to question #24.

**For the reaction below, K _{c} = 7.8×**

**10**^{8}. What is the equilibrium concentration of NH_{3}when 1.00 mol each of Zn(NO_{3})^{2}and NH_{3}are dissolved in water to make 1.0 L of solution?Zn^{2+} (aq) + 4 NH_{3} (aq) ⇌ Zn(NH_{3})_{4}^{2+}(aq)

This question puts a “twist” on the usual equilibrium problem because it involves a reaction with a very large equilibrium constant. Unfortunately, too much emphasis is placed on solving equilibrium problems for reactions having a small equilibrium constant. The most useful reactions typically have very large equilibrium constants and, for such reactions, it is helpful to be able to estimate how much of the limiting reactant remains unreacted. To solve this problem, it is important to recognize that because the equilibrium constant is large, the reaction goes almost (but not quite) to completion.

Consider the following equilibrium summary.

Reactants/Products | [Zn^{2}^{+}(aq)] |
[NH_{3} (aq)] |
[Zn(NH_{3})_{4}^{2+}(aq)] |
---|---|---|---|

Initial | 1.00 mol L^{−1} |
1.00 mol L^{−1} |
0 |

100% | 0.75 mol L^{−1} |
0 | 0.25 mol L^{−1} |

Change | + x | + 4x | − x |

Equilibrium | 0.75 + x | 4x | 0.25 − x |

The equilibrium summary has an additional line, which is labeled “100%”. The concentrations shown in this line are the concentrations expected if it is assumed the reaction goes to completion (i.e., until the limiting reactant, NH_{3}, is completely consumed). However, no reaction goes 100% to completion! Let x be the amount by which we've overshot the true equilibrium state. Because *K*_{c} is so large, we anticipate that the reaction will go nearly to completion and thus, x will be small. Thus, we substitute [Zn^{2+}] = 0.75 mol L^{−1}, [NH_{3}] = 4x and [Zn(NH_{3})_{4}^{2+}] = 0.25 mol L^{−1} into the equilibrium constant expression and solve for x. We get x = 0.001136 mol L^{-1} and and [NH_{3}] = 4x = 0.0045 mol L^{−1}. In other words, 99.55% of the NH_{3} reacted, leaving only 0.45% of the NH_{3} unreacted.

When solving this problem, many students will set up the following equilibrium summary.

Reactants/Products | [Zn^{2}^{+}(aq)] |
[NH_{3} (aq)] |
[Zn(NH_{3})_{4}^{2+} (aq)] |
---|---|---|---|

Initial | 1.00 mol L^{−1} |
1.00 mol L^{−1} |
0 |

Change | − x | − 4x | + x |

Equilibrium | 1.00 − x | 1.00 − 4x | x |

There is nothing incorrect about the setup above. However, the assumption that 4x will be small compared to 1.00 is NOT valid. Students need to recognize that most of the NH_{3} will react and thus, the quantity that is small is 1.00 − 4x. This is where a change of variable can help. If we let y = 1.00 − 4x, then x = 0.25(1.00 − y). When the setup above is recast in terms of y, we get:

Reactants/Products | [Zn^{2}^{+}(aq)] |
[NH_{3} (aq)] |
[Zn(NH_{3})_{4}^{2+} (aq)] |
---|---|---|---|

Equilibrium | 0.75 − 0.25y | y | 0.25 − 0.25y |

Because y is small, we can substitute [Zn^{2+}] = 0.75, [NH_{3}] = y and [Zn(NH_{3})_{4}^{2+}] = 0.25 into the equilibrium constant expression and solve for y. We obtain y = 0.0045 mol L^{−1}. Although solving the problem by making a change of variable is perfectly acceptable, it involves a lot of unnecessary algebra. Overall, I prefer the first approach because (1) it is algebraically simpler; (2) it starts with a stoichiometry calculation, a limiting reagent problem, which students should have already mastered; and (3) it requires students to consider the magnitude of the equilibrium constant before setting up and solving the problem.