Main content

## Weak acid and base equilibria

# Worked example: Finding the percent ionization of a weak acid

AP.Chem:

SAP‑9 (EU)

, SAP‑9.C (LO)

, SAP‑9.C.1 (EK)

, SAP‑9.C.2 (EK)

, SAP‑9.C.5 (EK)

## Video transcript

- [Instructor] Let's say we have a 0.20 Molar aqueous
solution of acidic acid. And our goal is to calculate the pH and the percent ionization. The Ka value for acidic acid is equal to 1.8 times
10 to the negative fifth at 25 degrees Celsius. First, we need to write out
the balanced equation showing the ionization of acidic acid. So acidic acid reacts with
water to form the hydronium ion, H3O+, and acetate, which is the
conjugate base to acidic acid. Because acidic acid is a weak acid, it only partially ionizes. Therefore, we need to set up an ICE table so we can figure out the equilibrium concentration
of hydronium ion, which will allow us to calculate the pH and the percent ionization. In an ICE table, the I stands
for initial concentration, C is for change in concentration, and E is equilibrium concentration. The initial concentration of
acidic acid is 0.20 Molar. So we can put that in our
ICE table under acidic acid. And if we assume that the
reaction hasn't happened yet, the initial concentrations
of hydronium ion and acetate anion would both be zero. Some of the acidic acid will ionize, but since we don't know how much, we're gonna call that x. So we write -x under acidic acid for the change part of our ICE table. And when acidic acid reacts with water, we form hydronium and acetate. So we're going to gain in
the amount of our products. To figure out how much
we look at mole ratios from the balanced equation. There's a one to one mole ratio of acidic acid to hydronium ion. Therefore, if we write -x for acidic acid, we're gonna write +x under hydronium. And for the acetate
anion, there's also a one as a coefficient in the balanced equation. Therefore, we can write
+x under acetate as well. So the equilibrium
concentration of acidic acid would be 0.20 minus x. Let's go ahead and write that in here, 0.20 minus x. The equilibrium concentration of hydronium would be zero plus x, which is just x. And for acetate, it would
also be zero plus x, so we can just write x here. Next, we brought out the
equilibrium constant expression, which we can get from
the balanced equation. So the Ka is equal to the concentration of the hydronium ion. And since there's a coefficient of one, that's the concentration of hydronium ion raised
to the first power, times the concentration
of the acetate anion also raised to the first power, divided by the concentration of acidic acid raised to the first power. And water is left out of our equilibrium constant expression. Because the concentrations in our equilibrium constant expression or equilibrium concentrations, we can plug in what we
have from our ICE table. So we can plug in x for the
equilibrium concentration of hydronium ion, x is also the equilibrium concentration of the acetate anion, and 0.20 minus x is the
equilibrium concentration of acidic acid. We also need to plug in the
Ka value for acidic acid at 25 degrees Celsius. Here we have our equilibrium
concentrations plugged in and also the Ka value. Our goal is to solve for x, which would give us the
equilibrium concentration of hydronium ions. However, if we solve for x here, we would need to use a quadratic equation. So to make the math a little bit easier, we're gonna use an approximation. We're gonna say that 0.20 minus x is approximately equal to 0.20. The reason why we can
make this approximation is because acidic acid is a weak acid, which we know from its Ka value. Ka is less than one. And that means it's only
going to partially ionize. It's going to ionize
to a very small extent, which means that x must
be a very small number. And if x is a really small
number compared to 0.20, 0.20 minus x is approximately
just equal to 0.20. So we can go ahead and rewrite this. So we would have 1.8 times
10 to the negative fifth is equal to x squared over, and instead of 0.20 minus x, we're just gonna write 0.20. Solving for x, we would
find that x is equal to 1.9, times 10 to the negative third. And remember, this is equal to
the equilibrium concentration of hydronium ions. So let's write in here, the equilibrium concentration
of hydronium ions. So this is 1.9 times 10 to
the negative third Molar. If we would have used the
quadratic equation to solve for x, we would have also gotten 1.9
times 10 to the negative third to two significant figures. Therefore, using the approximation
got us the same answer and saved us some time. Also, now that we have a value for x, we can go back to our approximation and see that x is very
small compared to 0.20. So 0.20 minus x is
approximately equal to 0.20. Also, this concentration of hydronium ion is only from the
ionization of acidic acid. And it's true that
there's some contribution of hydronium ion from the
autoionization of water. However, that concentration
is much smaller than this. So for this problem, we
can ignore the contribution of hydronium ions from the
autoionization of water. Next, we can find the pH of our solution at 25 degrees Celsius. So pH is equal to the negative
log of the concentration of hydronium ions. So that's the negative log of 1.9 times 10 to the negative third, which is equal to 2.72. We also need to calculate
the percent ionization. So the equation 4% ionization is equal to the equilibrium concentration
of hydronium ions, divided by the initial
concentration of the acid, times 100%. The equilibrium concentration
of hydronium ions is equal to 1.9 times 10
to negative third Molar. So we plug that in. And the initial concentration
of our weak acid, which was acidic acid is 0.20 Molar. So the Molars cancel, and we get a percent ionization of 0.95%. A low value for the percent
ionization makes sense because acidic acid is a weak acid. We can also use the percent
ionization to justify the approximation that
we made earlier using what's called the 5% rule. So let me write that
down here, the 5% rule. If the percent ionization is less than 5% as it was in our case, it
was less than 1% actually, then the approximation is valid. If the percent ionization
is greater than 5%, then the approximation is not valid and you have to use
the quadratic equation.

AP® is a registered trademark of the College Board, which has not reviewed this resource.