Kaplan MCAT General Chemistry Review

Keep in mind that K_b is nothing more than an equilibrium constant for the reaction of a base picking up a proton from water. So all of the things that are true for K_eq are true for K_b, especially that K_eq only depends on temperature. At constant temperature, it never changes (as the name suggests), even if the concentrations of the species in solution change.

Remember: The p-scale is a hugely important value for acid-base problems. Remember that p(something) = -log(something).

3) Write down the appropriate chemical reaction and set up a table.

Here we need to set up a table to reflect the data we’ve collected. This is the “putting it all together” step and is crucial.

Our table will be as follows:

H₂O (l ) + CH₃CO₂^-(aq) CH₃CO₂H (aq) + OH^-(aq) Initial—0.100Change—- x + x + x Equilibrium—0.1 - x x x

Initial: The idea is that we’re going to take some sodium acetate, dump it into water, and see what happens. Our initial row in the table shows the concentrations that we have before any reaction takes place. That means that we’ll start with the amount of sodium acetate we computed, 0.1 M. There’s no acetic acid or hydroxide because no reaction has happened yet.

Change: Here’s where all the action happens. As our acetate reacts with water, the concentration is going to decrease by some amount. We don’t know what that will be yet, so let’s just call it x. If the acetate concentration goes down by x, the concentrations of acetic acid and hydroxide must go up by the same amount, so we put x’s in their columns.

Equilibrium: This is the easy part. Just add up all of the columns above.

Remember: Always, always, always make sure that any chemical reaction you write down is balanced. This means to make sure that mass is balanced (the number of atoms on either side of the reaction) and that charge is balanced. Also, the concentration of pure liquids (e.g., water) and pure solids is never taken into account in equilibria.

Things to Watch Out For

Remember that the cardinal principle of handling computation on the MCAT is to avoid it whenever possible because it is so time consuming and drastically increases the chances of making a mistake. If you can’t avoid doing computation, choose numbers that are easy to work with. So, for example, you wouldn’t want to use 199.9999; you would just use 200.

4) Plug the equilibrium concentrations from the table into the appropriate acidity or basicity expression.

We know that K_b is just the K_eq for the reaction in our table above. Plug in the numbers from the table.

5) Simplify the expression from step 4 and solve it.

Let’s make our lives easier (and save ourselves time on Test Day) by assuming that x is much, much smaller than 0.1, so that 0.1 - x 0.1.

Now why would we want to assume that x is very, very small? Well, remember that sodium acetate is a weak base because it has a K_b value. So if it’s a weak base, it won’t react much with water, thus making x a very small number.

Remember: Don’t forget to check the assumption we made to simplify our equation. Because x = 10^-5, which is indeed much less than 0.1 (by a factor of 10,000), our assumption holds.

Similar Questions

1) What would be the pH if the initial concentration of sodium acetate in the opening question was halved? If it were doubled?

2) Sodium acetate is a buffer, a species that resists changes in pH. To see why this is true, compute the pH of the resulting solution if 0.1 mol of pure NaOH were added to 1 L of water. How does this pH compare to that of the solution with sodium acetate?

3) What would the pH of the solution be if just enough HCl were added to the solution in the original problem to consume all of the sodium acetate?

6) Answer the question.

-log[OH^-] = -log[10^-5] = 5 = pOH

pH = 14 - 5 = 9