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## AP®︎/College Chemistry

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

Lesson 3: Weak acid and base equilibria# Weak acid equilibria

AP.Chem:

SAP‑9 (EU)

, SAP‑9.C (LO)

, SAP‑9.C.1 (EK)

, SAP‑9.C.2 (EK)

, SAP‑9.C.5 (EK)

Unlike strong acids, weak acids only partially ionize in aqueous solution. As a result, the concentration of H₃O⁺ in a weak acid solution is typically much less than the initial acid concentration. The relative strength of a weak acid is described by the acid ionization constant,

*Kₐ*, which is the equilibrium constant for the reaction of the weak acid with water. Created by Jay.## Want to join the conversation?

- Why is pure water left out of the equation?(4 votes)
- It is conventional to omit pure liquids (like water), solids, and solvents from equilibrium expressions. This is a consequence of exactly how each quantity is defined. They aren't defined simply as the concentrations of the chemical species but rather as the ratio of the concentration of a species to its concentration in its standard state. This standard state is defined differently depending on the chemical species. For solutes, the standard state is 1 M, For gases the standard state is 1 bar, and for solids and liquids the standard states are the pure solid or liquid. Mathematically this means that the concentrations of a solid or liquid are the same as their concentration in the standard state creating a ratio equal to 1. This results in solid or liquid quantities essentially being canceled out since they simplify to 1. If it's a solvent then the concentration is so close to that of a pure liquid that the quantity is essentially 1 as well. This is actually also why equilibrium constants are unitless since defining all the quantities as such results in them being unitless making the entire expression dimensionless. This convention is arbitrary and there's no deeper meaning as to why we do it other than to make the math simpler.

Hope that helps.(10 votes)

- Since pH + pOH=14.00 is derived from the euqation: [H+]*[OH-]=K(water), why we can still apply pH + pOH = 14.00 to all aqueous solution even when K(acid)>>1, which is much larger than K(water). Wouldn't it be pH + pOH = pK(acid)(probably much smaller than 14.00)?(1 vote)
- The autoionization of water reaction, described by Kw, is a separate reaction from an acid dissociation reaction, described by Ka.

Ka>>1 means that the acid dissociation reaction lies far to the right, towards the products. It also means that the acid is strong and has a large majority of the original acid molecules dissociate. As opposed to a weak acid with a small Ka which only dissociates partially.

If we have a strong acid dissociate, then it contributes some hydronium to the aqueous solution to make it more acidic. But just as the acid reaction reaches an equilibrium, the water itself also reaches its own equilibrium. At 25°C Kw is 1.0x10^(-14) and that means the water autoionization reaction, Kw = [H3O^(+)][OH^(-)], always has to equal that number at that temperature. If we increase the hydronium concentration, then the hydroxide concentration decreases too to maintain that Kw value.

And since the Kw equation is equal to the pKw = 14 = pH + pOH equation, changes in the hydronium concentration cause subsequent changes in the pH and the pOH. But again since Kw is constant at 25°C, and pKw is based on Kw, the previous equation always equals 14. Strong acid or weak acid, doesn’t matter the Ka, pKw is 14.

Hope that helps.(3 votes)

- At the end of the video, it says that the % ionization of the hypothesized reaction is 20, but then when you do your calculations with the formula, you get 2.5%. I'm confused on which one is true.(1 vote)
- The 20% is for the hypothetical reaction using the particulate diagrams (the pictures with colored dots)10:00. The 2.5% is for an actual reaction using benzoic acid from the ICE table10:37.(1 vote)

## Video transcript

- [Instructor] Before
we get into the topic of weak acid equilibria, let's
look at a strong acid first. So let's say that HA is a strong
acid and reacts with water to produce the hydronium ion and A-minus, the conjugate base to HA. Technically, the reaction
comes to an equilibrium. However, the equilibrium
favors the products so much that we don't draw an equilibrium arrow. We simply draw an arrow going to the right indicating at equilibrium,
we have essentially all H30-plus and A-minus, and no HA. So we assume that a strong
acid ionizes 100% in solution, and we can see that in
our particulate diagrams. And keep in mind that
these particulate diagrams are not meant to represent
the entire solution, just a small portion of
it so we can get an idea about what's happening
in the entire solution. We can think about the
particulate diagram on the left representing the initial
concentration of our acid, HA. So there are five HA particles in solution and also five water particles. Since the reaction goes to completion, the HA particles are going to
react with the water particles to produce five hydronium,
H3O-plus, and five A-minus. We can see that in our
second particulate diagram, which represents the equilibrium mixture and there are five hydroniums present and also five A-minuses. We can write the equilibrium
constant expression for this reaction from
the balanced equation. we look at our products and
say it'd be the concentration of H3O-plus, and since
there's a coefficient of one, it'd be the concentration of H30-plus raised to the first power,
times the concentration of A-minus also raised to
the first power, divided by the concentration of HA
raised to the first power, and pure water is left out of the equilibrium constant expression. So that's equal to the
equilibrium constant K, and since this shows the
ionization of an acid, we could call this Ka value, the acid ionization constant, or the acid dissociation constant. Since the equilibrium
lies so far to the right for a strong acid, at
equilibrium, there's going to be essentially all products and
very, very little reactants. That means the Ka value for a strong acid will be much greater than one. Now let's look at a weak acid. We're gonna use the same
general equation here. HA plus H2O yields H3O-plus plus A-minus. However, if HA is a generic weak acid, we're going to include an
equilibrium arrow here. The equilibrium arrow needs to be included because weak acids only partially ionize, and we can see that in
the particulate diagrams. In the particulate diagram on the left, this is supposed to represent
the initial concentration of our weak acid HA and there
are five particles of HA and five particles of water. Since weak acids only
partially ionize, let's say that only one HA particle
reacts with one water molecule to produce one hydronium
ion and one A-minus anion. So if one HA particle reacts
with one H2O particle, that gives us one hydronium,
H30-plus and one A-minus. So out of the original five HA particles, only one of them ionized,
leaving four HA particles still in solution at
equilibrium, therefore HA only partially ionized and is a weak acid, the equilibrium constant
expression for a weak acid looks the same as it
did for our strong acid, because we use the same general equation. However, since a weak acid
only partially ionizes at equilibrium, there's
gonna be a small amount of products and a large
amount of reactants. Therefore, the Ka value for a weak acid will be less than one. Let's look at some Ka values for some weak acids at 25 degrees Celsius. Hydrofluoric acid has a Ka value of 6.3 times 10 to the negative fourth, and acetic acid has a Ka value of 1.8 times 10 to the negative fifth. Since both of these have
Ka values less than one, they're both considered to be weak acids. However, the higher the Ka
value, the stronger the acid. So out of these two weak
acids, since hydrofluoric acid has the higher Ka value, hydrofluoric acid is the stronger acid of the two. Now let's do a calculation
using a weak acid. Let's say we have a 0.10 molar
solution of benzoic acid, which is a weak acid, and
the pH of the solution is 2.60 at 25 degrees Celsius. Our goal is to calculate the Ka value for benzoic acid at 25 degrees Celsius. Since we're given the pH in the problem, we can go ahead and plug the
pH of 2.60 into this equation. So 2.60 is equal to the negative
log of the concentration of hydronium ions, and to
solve for the concentration of hydronium ions, first
moved the negative sign to the left, which gives us negative 2.60, is equal to log of the
concentration of hydronium ions. And to get rid of the log,
we take 10 to both sides. This gives us the
concentration of hydronium ions is equal to 2.5 times 10 to
the negative third molar. It's also important to recognize this is the equilibrium concentration
of hydronium ions. The next step is to write out the equation showing benzoic acid reacting with water. So here's the chemical
formula for benzoic acid reacting with water to
form the hydronium ion and the conjugate base to benzoic acid, which is the benzoate anion. To help us find the Ka
value for benzoic acid, we're gonna use an ICE table, where I stands for the
initial concentration, C is the change in concentration, and E is the equilibrium concentration. In our problem, we were told
the initial concentration of benzoic acid was 0.10
molar, and if we assume that benzoic acid hasn't reacted yet, the initial concentration of
hydronium ion would be zero, and the initial concentration of benzoate would also be zero. We just calculated the
equilibrium concentration of hydronium ions, which we found to be 2.5 times 10 to the negative third molar. So we're going to go
ahead and put that under our hydronium ion in for the
equilibrium concentration. So if the initial concentration
of hydronium ion was zero and the equilibrium concentration is 2.5 times 10 to the
negative third, there was an increase in 2.5 times
10 to the negative third, so we're going to write
that under hydronium for the change in the concentration. And looking at the mole
ratio of hydronium ion to benzoate, it's a one-to-one mole ratio. So if hydronium ion increased by 2.5 times 10 to the negative third, so did the benzoate anions,
I can go ahead and write here plus 2.5 times 10 to the negative third, which means the equilibrium concentration of benzoate anion would also be 2.5 times 10 to the negative third. Since we saw an increase
in the concentration of hydronium ion and
benzoate, that increase in the concentration of the products came from the ionization
of our weak acids. In the balanced equation, there's a one in front of benzoic acid, so if we write plus 2.5 times 10 to the
negative third for hydronium, and it's a one-to-one
mole ratio from hydronium to benzoic acid, we would have to write minus 2.5 times 10 to the negative third, since we lost some of the
benzoic acid when it ionized, therefore the equilibrium
concentration of benzoic acid would be 0.10 minus 2.5 times
10 to the negative third. Since our goal is to
calculate the Ka value for benzoic acid, the
next step is to write out the equilibrium constant expression, which we get from the balanced equation. It would be the concentration
of hydronium ions times the concentration of benzoate anions divided by the concentration
of benzoic acid. Since these are
equilibrium concentrations, we can get the equilibrium concentrations from the ICE table and plug them in to our equilibrium constant expression. So we can plug in the
equilibrium concentration for hydronium ions, we can plug in the equilibrium concentration
for benzoate anions, and we can plug in the
equilibrium concentration for benzoic acid. Here we have the
equilibrium concentrations plugged in to the equilibrium
constant expression, and when we solve for Ka, we find that Ka for benzoic acid at 25 degrees Celsius is equal to 6.4 times 10
to the negative fifth. Notice that Ka is less than one because benzoic acid is a weak acid. Finally, let's talk about the
idea of percent ionization. And to help us think
about percent ionization. Let's go back to our
example with a strong acid. We started off with five HA particles and all five of those HA particles ionized to produce five hydronium ions in solution and five A-minus anions in solution. Therefore 100% of our acid ionized. So we can write 100% ionization for this hypothetical strong acid HA. Now let's go back to our weak acid. For our weak acid, we started
off with five HA particles, but only one of them ionized to produce one H3O-plus and one A-minus. So four remain unionized at equilibrium, therefore one out of five
ionized, so the percent ionization for this hypothetical
weak acid would be 20%. Notice that the one
particle of HA that ionized made one hydronium ion, so we can also use the concentration of
hydronium ion at equilibrium to calculate percent ionization. So the equation for percent Ionization is equal to the equilibrium concentration of hydronium ions divided
by the initial concentration of the acid HA times 100%. So looking at our ICE table
for our benzoic acid problem, the equilibrium concentration
of hydronium ions was 2.5 times 10 to the
negative third molar, and the initial concentration
of our acid was 0.10 molar, so we can plug those into our equation. So we plug in our concentrations
and molar cancels, and we find for the final
answer, the percent ionization of benzoic acid is equal to 2.5%.