Kaplan MCAT General Chemistry Review

This step is trickier than it sounds and is where many, many mistakes are committed. Here’s where attention to detail counts. You don’t want to slog through all of the work above and then mess up at the end, when 99 percent of the work is done!

Think about what you’ve solved for. What is x? Well, if we look at the table, we see that x is the concentration of hydroxide. So if we take the negative log of the hydroxide ion concentration, we get the pOH.

Remember that pH + pOH = 14. The actual pH is 8.89. So all of our assumptions and roundings didn’t affect the answer much, but they saved us a lot of time in computation!

Remember: Always ask yourself whether your final answer makes sense. The MCAT isn’t a computation test; it’s a test of critical thinking. Here, we have a base being dissolved in water, so at the end of the day, the pH better be above 7, which it is.

Titration

Key Concepts

Chapter 10

Titration

Acids and bases

Equivalence point

Half-equivalence point

pH

Hydrazoic acid, HN₃, is a highly toxic compound that can cause death in minutes if inhaled in concentrated form. 100 mL of 0.2 M aqueous solution of HN₃ (pK_a = 4.72) is to be titrated with a 0.5 M solution of NaOH.

a) What is the pH of the HN₃ solution before any NaOH is added?

b) The half-equivalence point of a titration is where half the titrant necessary to get to the equivalence point has been added. How much of the NaOH solution will be needed to get to the half-equivalence point? What is the pH at the half-equivalence point?

c) What is the pH at the equivalence point?

1) Determine the pH before the titration.

What you need to ask yourself in each stage of this problem is which species is present, H⁺ or OH^-, and where is it coming from? Before the titration begins, we have H⁺ around because, as the name of the compound suggests, hydrazoic acid is acidic. The major source of H⁺ is from the hydrazoic acid itself, so we can set up a table as follows:

Takeaways

Setting up the tables as shown makes quick work of titration pH questions. Remember to make approximations and use numbers that are easy to work with in order to minimize the computation necessary to get to the answer.

Plug in the concentrations from the table. We know the pK_a is around 5, so the K_a must be 10^-5. Whenever exponents or logarithms are involved, use numbers that are easy to work with. Make the approximation that 0.2 - x 0.2 because HN₃ is a weak acid.

If you must take a square root, try to get the power of 10 to be even to make matters simple. Remember from the table that x is the hydronium ion concentration.

Remember that -log[a × 10^-b] = b - log[a] = somewhere between b - 1 and b. 2 < pH < 3

2) Find the equivalence point and half-equivalence point.

Compute the number of moles of HN₃ that you start with.

Each mole of HN₃ will react with one mole of OH^-. Therefore, the equivalence point is reached when 40 mL of the NaOH solution are added and the half-equivalence point is at 40/2 = 20 mL. We could go through the whole rigamarole of setting up another table to figure out the pH at the half-equivalence point, or we could use a little common sense to avoid computation. At the half-equivalence point, half of the HN₃ has been consumed and converted to N₃^-. Therefore, the HN₃ and N₃ concentrations are equal.

Because [HN₃] = [N₃^-], K_a = [H₃O⁺], and pH = pK_a = 4.72.

Remember: Whenever possible, avoid computation!

Things to Watch Out For

Be careful in choosing whether you will use K_a or K_b to determine the pH. Make this decision based on whether the dominant species in solution is acidic or basic, respectively.

3) Determine the reactive species at the equivalence point to find the pH.

At the equivalence point, all of the HN₃ has been consumed, leaving only N₃^- behind. Because N₃^- is a Brønsted-Lowry base, we need to worry about OH^-, not H₃O⁺.

Remember that K_a × K_b = K_w = 10^-14. Now we can set up our table:

Remember that the volume of our solution has increased by 40 mL, so the concentration of N ^{- 3} is 0.02 mol (0.1 + 0.04) L 0.15 M.

Make the approximation that 0.15 - x 0.15.

Similar Questions

1) What is the pH after the equivalence point has been exceeded by 5 mL of the NaOH solution in the opening question?