Worked example: Calculating the pH after a weak acid–strong base reaction (excess acid)

0:00- [Instructor] Let's look at a reaction between a weak acid, acetic acid, and a strong base, sodium hydroxide. Let's say we have 100 milliliters of a 2.0 molar solution of aqueous acetic acid, and that's mixed with 100 milliliters of a 1.0 molar solution of aqueous sodium hydroxide.

0:19Our goal is to find the pH of the resulting solution at 25 degrees Celsius. When the weak acid reacts with the strong base, a neutralization reaction occurs. So our first step is to figure out how many moles of our weak acid are present and also how many moles of our strong base are present.

0:37Let's start with the weak acid. We're gonna use the molarity equation. Molarity is equal to moles over liters. So for our weak acid, the concentration is 2.0 molar, so we plug that in, and the volume is 100 milliliters which is equal to 0.100 liters. Solving for x, we find that there are 0.20 moles of acetic acid.

0:59For our strong base, the concentration is 1.0 molar, and the volume is 100 milliliters which is 0.100 liters. So solving for x, x is equal to 0.10 moles of NaOH. And because NaOH is a strong base, it dissociates 100%. So if there's 0.10 moles of NaOH, there's also 0.10 moles of hydroxide ions, OH-.

1:26In solution, the hydroxide anions will react with acetic acid. So our next step is to look at the net ionic equation for this weak acid-strong base reaction. In the net ionic equation, acetic acid reacts with hydroxide anions to form the acetate anion and water.

1:44Because this neutralization reaction goes to completion, we draw an arrow going to the right instead of an equilibrium arrow. And instead of an ICE table where the E stands for equilibrium, we use an ICF table, where I is the initial amount of moles, C is the change in moles, and F is the final amount of moles.

2:02We've already calculated the initial amount of moles of acetic acid to be 0.20, and the initial amount of moles of hydroxide anions to be 0.10. If we assume the reaction hasn't happened yet, the initial amount of moles of acetate would be zero. Next, we look at the coefficients in our balanced net ionic equation.

2:22The mole ratio of hydroxide anions to acetic acid is one-to-one. Therefore, if we have 0.10 moles of hydroxide anions, that's gonna react with 0.10 moles of acetic acid. So for the change, we can write minus 0.10 under hydroxide anions, and also minus 0.10 under acetic acid.

2:46For the acetate anion, there's a coefficient of one in the balanced net ionic equation. Therefore, if we're losing 0.10 moles for our two reactants, we're going to gain 0.10 moles of the acetate anion. So when the reaction comes to completion for acetic acid, we started with 0.20 moles and we lost 0.10, so we're gonna have 0.10 moles of acetic acid left over.

3:11For hydroxide anions, we start with 0.10 moles and we lost 0.10 moles. Therefore, when the reaction comes to completion, there'll be zero moles of hydroxide anions. And for the acetate anion, we started off with zero and we gained 0.10. Therefore, we will have 0.10 moles of the acetate anion.

3:31So in this case, we started with more of our weak acid than we did of our strong base. And therefore, we ended up with acid being in excess. The next step is to calculate the concentration of acetic acid in solution. So we ended up with 0.10 moles of acetic acid.

3:48So we plug that into our equation for molarity. And when we mixed our two solutions together, the 100 milliliters of our weak acid with the 100 milliliters of our strong base, the total volume of the solution is 200 milliliters or 0.200 liters. Therefore, 0.1 divided by 0.200, gives the concentration of acetic of 0.50 molar.

4:12And for the acetate anion, we also had 0.10 moles. The total volume is the same, so it's the same calculation as before. 0.10 moles divided by 0.200 liters, gives the concentration of acetate anions of 0.50 molar. Remember, our goal was to calculate the pH of the resulting solution.

4:32So to calculate the pH, we're gonna need the concentration of acetic acid and the concentration of acetate anions. So we know we have some acetic acid in solution, and acetic acid reacts with water to form the hydronium ion, H3O+, and the acetate anion, CH3COO-.

4:51This reaction is a weak acid equilibrium problem, so we have an equilibrium arrow, and we're gonna set up an ICE table for initial concentration, change in concentration, and equilibrium concentration. We just calculated the initial concentration of acetic acid to be 0.50 molar.

5:08And if we pretend like the acetic acid hasn't ionized yet, the initial concentration of hydronium ions would be zero and the initial concentration of acetate anions from the ionization would be zero. However, we just calculated that there is a concentration of acetate anions already in solution, and that concentration was 0.50 molar.

5:30So on the ICE table, I'm gonna write in here 0.50 for the acetate anion. Next, we think about some of the acetic acid ionizing, and we don't know how much, so we're gonna call that x. So under the change, we're gonna write minus x. The mole ratio of acetic acid to hydronium ion is one-to-one.

5:50So if we write minus x under acetic acid, we need to write plus x under the hydronium ion. And for the acetate anion, there's also a coefficient of one, so we're gonna write plus x under the acetate anion. Therefore, the equilibrium concentration of acetic acid is 0.50 minus x.

6:09And for hydronium ion, it would be zero plus x or just x. And for the acetate anion, it would be zero plus 0.50 plus x or just 0.50 plus x. Notice that there are two sources of the acetate anion. The 0.50 molar came from the weak acid-strong base neutralization reaction that we previously discussed.

6:35The other source of the acetate anion comes from the ionization of acetic acid, and that's this x here. So 0.50 plus x indicates there are two sources for this common ion. Therefore, this is a common-ion effect problem. The next step is to write the Ka expression for acetic acid.

6:56So we can get that from the balanced equation. The Ka value for acetic acid at 25 degrees Celsius is equal to the concentration of hydronium ions raised to the first power times the concentration of acetate anions raised to the first power divided by the concentration of acetic acid raised to the first power, with water being left out of our equilibrium constant expression.

7:18Next, we plug in our equilibrium concentrations from our ICE table. For the hydronium ion, it's x. For the acetate anion, it's 0.50 plus x. And for acetic acid it's 0.50 minus x. Here, we have our equilibrium concentrations plugged in and also the Ka value for acetic acid at 25 degrees Celsius.

7:40Next, we need to solve for x. And to make the math easier, we can make an approximation. With a relatively low value for Ka for acetic acid, acetic acid doesn't ionize very much. Therefore, we know that x is gonna be a very small number. And if x is a very small number compared to 0.50, 0.50 plus x is approximately equal to 0.50.

8:03The same idea applies to 0.50 minus x. If x is a very small number compared to 0.50, 0.50 minus x is approximately equal to 0.50. And with our approximations, the 0.50s would cancel out and give us x is equal to 1.8 times 10 to the negative fifth. From our ICE table, we know that x is equal to the equilibrium concentration of hydronium ions.

8:27Therefore, at equilibrium, the concentration of hydronium ions is equal to 1.8 times 10 to the negative fifth molar. And because our goal was to find the pH of the solution, pH is equal to the negative log of the concentration of hydronium ions. So when we plug in that concentration, we get the pH is equal to negative log of 1.8 times 10 to the negative fifth which is equal to 4.74.

8:52So if you react a weak acid with a strong base and the weak acid is in excess, the pH of the solution will be less than seven. It'll be acidic.