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### Course: AP®︎/College Chemistry>Unit 8

Lesson 5: Acid–base reactions

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

When a strong base reacts with excess weak acid, the resulting solution contains both HA and A⁻. We can use reaction stoichiometry to determine the concentration of each species and then solve a common-ion equilibrium problem to find [H₃O⁺] and thus the pH. Created by Jay.

## Want to join the conversation?

• for this example, we have the same concentration of CH3COOH and CH3COO-, then why the reaction of CH3COO- + H2O is not considered but the reaction of CH3COOH + H2O is?
• Jonathan was correct:
CH3COO- (as a conjugate base of CH3COOH) is a lot less reactive than CH3COOH is as an weak acid. Which makes sense, since if conjugates aren't less reactive than the "original", then we would break the second law of thermodynamics and achieve perpetual motion by them continually reacting in opposite direction at significantly different rates and never reaching equilibrium.

Another way of thinking about it is:
the equilibrium constant of the reaction
CH3COOH + H2O <=> H3O+ + CH3COO
which was given to be k = 1.8 * 10^-5
already takes into account of the excess amount of CH3COO- present as a result of the previous reaction. However it dictates that despite the conjugate acid present, in order for the concentrations to be balanced out and the equilibrium to be reached, more CH3COOH still have to react with H2O. Therefore we are able to determine the direction of the reaction by calculating using the equilibrium constant alone (comparing the reaction quotient (initial concentration ratio) and the equilibrium constant)
• At , couldn't you just skip the rest of the problem since it's the half-equivalence point? Correct me if I'm wrong, but since the concentrations of CH3COOH and CH3COO- are equal, we should be able to say that pH = pKa = -log(1.8*10^-5) = 4.74.

Thanks for the help!
• You're right. However this being an explanatory video, Jay intended to communicate a conceptual understanding of what's going on, and directly using pKa defeats the purpose. In exams and stuff, you can feel free to use this shortcut (or the H-H equation if the concentrations aren't equal)
• Why are you using an ICE table for this problem when in the previous video you explicitly said that "If weak acid is excess use hendersson-hasselbalch equation."
• The Henderson-Hasselbalch equation is a shortened ICE table equilibrium problem. The equation assumes that a weak acid on its conjugate base are present in significant amounts in solution and that they reach equilibrium (thus creating a buffer solution). It always assumes the ‘x is small approximation’ is valid, or the change in concentrations for the acid and base from their initial concentrations to their equilibrium concentrations are so small that it is insignificant and can be omitted.

The Henderson-Hasselbalch equation takes as inputs the concentrations of the acid and base to determine the pH of the solution (along with the pKa). In this problem we don’t have that; instead we are mixing a weak acid, acetic acid, with a strong base, sodium hydroxide. So we don’t know the concentrate of the conjugate base, acetate, because isn’t present initially.

We first have to determine the moles of acetate formed from the neutralization reaction of acetic acid and sodium hydroxide. Once we know the moles of acetate formed, and the remaining moles of acetic acid, we can calculate their molarities. Only then could we use the Henderson-Hasselbalch equation. Jay instead here chooses to use an ICE table to calculate the pH, but it still results in the same answer had we used Henderson-Hasselbalch equation.

Hope that helps.
• What do you do if there is no excess reactant because both the acid and base have the same concentration and molar ratio, so both end up being zero and you only have the products left. I'm stuck at the ICF table.
• Why is it that for these problems, we have been using the number of moles when completing the ICF table as opposed to the concentration for the ICE table?
(1 vote)
• Since these are acid-base reactions, we need an ICF table to know how many moles of the existing buffer acid and base are neutralized and turned into the conjugate by the added acid or base. This added acid or base also increases the volume of the solution changing the molarity of the buffer acid and base. So once we have the correct new molarities from the ICF table, they can be inputted into the ICE table which deals with equilibrium and molarities.

Hope that helps.
• does it have to be Ka in or is there a way you can make it Kb?
• This is an acid dissolution reaction so there's no point in using Kb
(1 vote)
• Are we already supposed to know the Ka value for acetic acid?
(1 vote)
• No, we aren't supposed to know the Ka, it should be given to us. It's just that Jay didn't give us Ka at the start of the question.
• why we do not use equilibrium arrow in the first equation but we use in the second equation?
(1 vote)
• how do we know if there will be excess acid? (AKA when do I know to solve the problem in this way?)
(1 vote)
• It’s essentially the same process of determining a limiting reactant. The other non-limiting reactant will automatically be in excess.

You need to be able to convert the volumes and molarities into moles of acid and base first. We also need the balanced chemical equation of the neutralization reaction. From equation we know how many moles of acid and base are consumed for a successful reaction. In this reaction between acetic acid and sodium hydroxide, they react in a 1:1 ratio meaning every mole of acetic acid needs a mole of sodium hydroxide to react. We know there are more moles of acetic acid than hydroxide, so essentially we are limited by how much sodium hydroxide we have and therefore will have an excess of acetic acid.

Hope that helps.
(1 vote)

## Video transcript

- [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. Our 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. Let'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. For 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-. In 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. Because 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. We'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. The 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. For 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. For 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. So 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. So 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. And 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. So 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-. This 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. And 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. So 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. So 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. And 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. The 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. So 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. Next, 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. Next, 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. The 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. Therefore, 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. So 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.