Kaplan MCAT General Chemistry Review

Be careful when applying Le Châtelier’s principle in cases of precipitation and solvation. For a solution at equilibrium (i.e., saturated), adding more solid would not shift the equilibrium to the right. More solid does not dissociate to raise the ion concentrations; the solid just piles up at the bottom.

3) Calculate molar solubility of each product by assuming that you’re starting withxmols of reactant.

x mol Fe(OH)₃x mol Fe ⁺³ + 3x mol OH^-

Use the balanced equation from step 1 to determine the appropriate coefficients. For each mole of iron hydroxide dissolved, four ions are created: one Fe⁺³ and three OH^-.

4) Plug the molar solubility for each product into the K_spequation:

Fe(OH)₃: K_sp = [Fe⁺³] [OH^-]³

K_sp = [x] [3x]³ = 27x⁴ (x = 9.94 × 10^-10)

K_sp = 27(9.94 × 10^-10)⁴ = 2.64 × 10^-35

Simply plug in the coefficients from step 3 into the K_sp equation. The molar solubility, x, was given in the question stem. Thus, to calculate K_sp, plug in 9.94 × 10^-10. Now that we are armed with the K_sp, we can find the change in molar solubility due to the common ion.

5) If a common ion is present in a solution, it must also be accounted for in the K_spequation:

Fe(OH)₃: K_sp = [Fe⁺³] [OH^-]³

2.64 × 10^-35 = x(3x + 10^-4)³x(10^-4)³

2.64 × 10^-35 = x10^-12

x = 2.64 × 10^-23

A solution of pH 10.0 has an OH^- concentration of 10^-4. Although iron hydroxide will also contribute to the solution’s total concentration of OH^–, its contribution will be negligible relative to the 10^-4 already present in solution; thus, when plugging in for the [OH^-], we can approximate that it equals 10^-4. For the pH 10.0 solution, the molar solubility is on the order of 10^-23 mol/L, a steep decrease from the 10^-9 in pure water. This is due to the common ion effect. Look at it from the perspective of Le Châtelier’s principle: The addition of more OH^- will shift the reaction to the left, so less iron hydroxide will dissociate.

Similar Questions

1) Given a substance’s K_sp, how would you solve for its molar solubility in pure water? What if a common ion were also present in solution?

2) Given a table listing substances and their solubility constants, how would you determine which substance was most soluble in pure water?

3) Given that the sulfate ion can react with acid to form hydrogen sulfate, how would the molar solubility of sulfate salts be affected by varying a solution’s pH?

pH and pK_a

Key Concepts

Chapter 10

Acids and bases

Chemical equilibrium

What is the pH of the resulting solution if 4 g of sodium acetate (CH₃CO₂Na) is dissolved in 0.5 L of water? (The pK_a of acetic acid, CH₃CO₂H, is 4.74.)

1) Convert masses to concentrations.

Molecular weight of sodium acetate in g mol^-1: 23 + 2(12) + 3(1) + 2(16) = 82. The concentration of a solution is usually expressed in units of moles per liter.

Choose numbers that are easy to work with (i.e., 80 g mol^-1 instead of 82 g mol^-1). Remember, you won’t have a calculator on Test Day! Moles of sodium acetate:

Remember: Moles, not grams, are the “common currency” of chemistry problems dealing with reactions. Concentration, usually in units of moles per liter, is the essential quantity for these acid-base problems.

2) Choose the constant that will make most sense for the reaction in question and compute its value.

Now we need to decide which constant we will use, K_a or K_b. Here’s where a little common sense goes a long way. If you think about sodium acetate, you should realize that it is the conjugate base of acetic acid. Therefore, we need the value of K_b for sodium acetate. Even though we are given the pK_a of acetic acid, getting the K_b from this information is no sweat.

In water, K_a × K_b is always equal to K_w, or 10^-14. Take the negative logarithm of both sides.

-log(K_a × K_b) = - log(10^-14)

Remember that log(a × b) = log a + log b and that log 10^x = x.

-log K_a + -log K_b = - log 10^-14 = 14

pK_a + pK_b = 14

Don’t forget that -log(whatever) = p(whatever). We will want to find K_b because sodium acetate is the conjugate base of acetic acid.

Plug in the pK_a from above (to make our lives easier, let’s say 4.74 5).

pK_b = 14 - pK_a = 9

K_b = 10^-pKb = 10^-9

Takeaways