Worked example: Finding the percent ionization of a weak acid | AP Chemistry | Khan Academy

Let's say we have a 0.20 molar aqueous solution of acetic acid, and our goal is to calculate the pH and the percent ionization. The Ka value for acetic 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 acetic acid. So, acetic acid reacts with water to form the hydronium ion (H₃O⁺) and acetate, which is the conjugate base to acetic acid. Because acetic 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 the 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 for equilibrium concentration. The initial concentration of acetic acid is 0.20 molar, so we can put that in our ICE table under acetic acid. 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 acetic acid will ionize, but since we don't know how much, we're going to call that x. So we write minus x under acetic acid for the change part of our ICE table. When acetic 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 the mole ratios from the balanced equation. There's a one-to-one mole ratio of acetic acid to hydronium ion. Therefore, if we write minus x for acetic acid, we're going to write plus x under hydronium. For the acetate anion, there's also a one as a coefficient in the balanced equation; therefore, we can write plus x under acetate as well.

So, the equilibrium concentration of acetic 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 0 plus x, which is just x, and for acetate, it would also be 0 plus x, so we can just write x here.

Next, we write 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 (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 acetic acid raised to the first power. Water is left out of our equilibrium constant expression because the concentrations in our equilibrium constant expression are 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 acetic acid. We also need to plug in the Ka value for acetic 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 going to use an approximation. We're going to say that 0.20 minus x is approximately equal to 0.20.

The reason why we can make this approximation is because acetic 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. 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. 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 going to write 0.20. Solving for x, we would find that x is equal to 1.9 times 10 to the negative third. Remember, this is equal to the equilibrium concentration of hydronium ions.

So, let's write in here the equilibrium concentration of hydronium ions: 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 acetic 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 for percent ionization is equal to the equilibrium concentration of hydronium ions divided by the initial concentration of the acid times 100 percent. The equilibrium concentration of hydronium ions is equal to 1.9 times 10 to the negative third molar, so we plug that in, and the initial concentration of our weak acid, which was acetic acid, is 0.20 molar. So the molars cancel, and we get a percent ionization of 0.95 percent.

A low value for the percent ionization makes sense because acetic 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 five percent rule. So, let me write that down here: the five percent rule. If the percent ionization is less than five percent, as it was in our case (it was less than one percent, actually), then the approximation is valid. If the percent ionization is greater than five percent, then the approximation is not valid, and you have to use the quadratic equation.