Kaplan MCAT General Chemistry Review

2) What would be the pH if the same amount of NaOH solution necessary to get to the half-equivalence point in this titration were added to pure water? How does the pH of each situation compare? This demonstrates how weak acids can serve as buffers, solutions that resist changes in pH.

3) What would be the pH of a solution that was 0.2 M in HN₃ and 0.10 M in N₃^–?

Here, let’s say that 1.6 is close to 1.5 to make the square root computation trivial.

Note that pH + pOH = 14 in water. Remember to ask yourself whether or not a result makes sense. Here, because we have a basic species (N₃^-), the pH should be above 7, which it is.

Balancing Redox Reactions

Key Concepts

Chapter 11

Oxidation

Reduction

Balancing electrochemical half-reactions

Balance the following reaction that takes place in basic solution.

ZrO(OH)₂ (s) + SO₃^2- (aq) Zr (s) + SO₄^2- (aq)

1) Separate the overall reaction into two half-reactions.

ZrO(OH)₂Zr

SO₃^2-SO₄^2-

Break the reactions up by looking at atoms other than hydrogen and oxygen.

2) Balance the oxygens in each reaction by adding the necessary number of moles of water to the appropriate side.

ZrO(OH)₂Zr (s) + 3 H₂O

H₂O + SO₃^2-SO₄^2-

3) Balance hydrogen by adding the necessary number of H⁺ions to the appropriate side of each reaction.

4 H⁺ + ZrO(OH)₂Zr + 3 H₂O

H₂O + SO ₃^2-SO₄^2- + 2 H⁺

Takeaways

Don’t fall into the trap of simply balancing mass in these reactions. If oxidation and reduction are occurring, you must go through this procedure to balance the reaction.

4) If the reaction is carried out in basic solution, “neutralize” each equivalent of H⁺with one equivalent of OH^-.

4 OH^- + 4 H⁺ + ZrO(OH)₂Zr + 3 H₂O + 4 OH^-

2 OH^- + H₂O + SO₃^2- SO₄^2- + SO₄^2- + 2 H⁺ + 2 OH^-

4 H₂O + ZrO(OH)₂Zr + 3 H₂O + 4 OH^-

2 OH^- + H₂O + SO₃^2-SO₄^2- + 2 H₂O

Combine each mole of H⁺ and OH^- into one mole of water and simplify each reaction.

Remember: Don’t forget to add OH^-to each side of both reactions!

Things to Watch Out For

These kinds of problems can be extremely tedious. You must take extra care to avoid careless addition and subtraction errors!

5) Balance the overall charge in each reaction using electrons.

4 e^- + 4 H₂O + ZrO(OH)₂ Zr + 3 H₂O + 4 OH^-

2 OH^- + H₂O + SO₃^2- SO₄^2- + 2 H₂O + 2 e^-

The top equation has a total charge of -4 on the right from the 4 moles of hydroxide, so 4 electrons need to be added to the left side of the equation.

In the bottom equation, there is a total charge of -4 on the left, -2 from the 2 moles of hydroxide and -2 from the 2 moles of sulfite anion (SO₃^2-).

Remember: Don’t forget to account for all charges in this step, including the charge contributed by molecules other than H⁺and OH^-.

6) Multiply each reaction by the necessary integer to ensure that equal numbers of electrons are present in each reaction.

4 e^- + 4 H₂O + ZrO(OH)₂ Zr + 3 H₂O + 4 OH^-

4 OH^- + 2 H₂O + 2 SO₃^2- 2 SO₄^2- + 4 H₂O + 4 e^-

Here, the lowest common multiple among the four electrons in the top reaction and the two in the bottom is four electrons, so we must multiply everything in the bottom reaction by two.

7) Combine both reactions and simplify by eliminating redundant molecules on each side of the reaction.

4 e^- + 4 H₂O (l ) + ZrO(OH)₂ (s) + 4 OH^- (aq) + 2 H₂O (l ) + 2 SO₃^2- (aq)

Zr (s) + 3 H₂O (l) + 4 OH^- (aq) + 2 SO₄^2- (aq) + 4 H₂O (l) + 4 e^-

Combine common terms on each side of the net reaction.

4 e^- + 6 H₂O (l) + ZrO(OH)₂ (s) + 4 OH^- (aq) + 2 SO₃^2- (aq)

Zr (s) + 7 H₂O (l) + 4 OH^- (aq) + 2 SO₄^2- (aq) + 4 e^-

Eliminate the redundant water molecules, as well as the electrons and excess hydroxide equivalents.

ZrO(OH)₂ (s) + 2 SO₃^2- (aq) Zr (s) + 2 SO₄^2- (aq) + H₂O (l)

Check to make sure that the reaction is balanced, in terms of both mass (number of atoms on each side) and overall charge.