Here, we’ll show you how to calculate the pH and % ionization of a weak acid with a given concentration and a known Ka value. K a to pH and Percent Ionization

For the (a) part of this question, we’re asked to find the pH of a 0.25 M solution of ethanoic acid, CH3COOH. a) Find the pH of 0.25 M CH 3 COOH

The “b” part of the question asks us to find the % ionization in 0.25 M ethanoic acid. a) Find the pH of 0.25 M CH 3 COOH b) Find the % ionization of 0.25 M CH 3 COOH

We’ll start with the “a” part. We’re asked to find the pH of a 0.25 M solution of ethanoic acid, CH3COOH. a) Find the pH of 0.25 M CH 3 COOH

Whenever we’re asked to do a calculation with an acid, the first thing we always have to do is identify the acid as strong or weak. a) Find the pH of 0.25 M CH 3 COOH Strong Acid or Weak Acid?

We see by its position on the acid table (click) that ethanoic acid is a weak acid a) Find the pH of 0.25 M CH 3 COOH Strong Acid or Weak Acid? Weak Acid

When we’re doing calculations involving weak acids, we must use an ICE table. a) Find the pH of 0.25 M CH 3 COOH Strong Acid or Weak Acid? Weak Acid ICE Table

We set up an ice table like this a) Find the pH of 0.25 M CH 3 COOH [I] [C] [E]

We start by writing the equilibrium equation for ionization of ethanoic acid here. a) Find the pH of 0.25 M CH 3 COOH [I] [C] [E] Equilibrium Equation

Next, we draw borders in so that columns line up nicely with the substances in the equation. a) Find the pH of 0.25 M CH 3 COOH [I] [C] [E] Equilibrium Equation

Water is a liquid in the equilibrium equation so we can ignore the column below water. We’ll colour it blue here. a) Find the pH of 0.25 M CH 3 COOH [I] [C] [E]

We’ll start out with the initial concentration row. a) Find the pH of 0.25 M CH 3 COOH [I]1 [C]1 [E]1

The initial concentration of the ethanoic acid is 0.25 M, so will add 0.25 to this cell. a) Find the pH of 0.25 M CH 3 COOH [I]0.251 [C]1 [E]1

No H3O+ or CH3COOminus was added, so we can consider their concentrations to be 0 before ionization. a) Find the pH of 0.25 M CH 3 COOH [I] [C]1 [E]1

Now we’ll look at the (click) changes in concentration as the ionization occurs. a) Find the pH of 0.25 M CH 3 COOH [I] [C]1 [E]1

Because there were no products initially, the equilibrium will (click) shift to the right during the ionization. a) Find the pH of 0.25 M CH 3 COOH [I] [C]1 [E]1 Shift to the Right

As a shift to the right occurs, the concentrations of hydronium and ethanoate ions will both increase so we’ll write + signs here. a) Find the pH of 0.25 M CH 3 COOH [I] [C]1+?+?+?+? [E]1 Shift to the Right

And the concentration of CH3COOH will decrease, so we’ll write a minus sign here. a) Find the pH of 0.25 M CH 3 COOH [I] [C]–x–x1+?+?+?+? [E]1 Shift to the Right

We’re not given any equilibrium concentrations, so we don’t know how much these will increase a) Find the pH of 0.25 M CH 3 COOH [I] [C]–x–x1+?+?+?+? [E]1 Shift to the Right

So we’ll write + x for both of them, as they both have a coefficient of 1 in the equilibrium equation. a) Find the pH of 0.25 M CH 3 COOH [I] [C]–x–x1+x+x+x+x [E]1 Shift to the Right

Because the coefficient on CH3COOH is also 1, we can state that it will go down by x, so we write minus x here. a) Find the pH of 0.25 M CH 3 COOH [I] [C]–x–x1+x+x+x+x [E]1 Shift to the Right

Now, we’ll look at the concentrations of everything (click) at equilibrium. a) Find the pH of 0.25 M CH 3 COOH [I] [C]–x–x1+x+x+x+x [E]1

The hydronium and ethanoate ions will both be 0 a) Find the pH of 0.25 M CH 3 COOH [I] [C]–x–x1+x+x+x+x [E] 0.25–x 1xx

Plus x a) Find the pH of 0.25 M CH 3 COOH [I] [C]–x–x1+x+x+x+x [E] 0.25–x 1xx

Which is equal to a) Find the pH of 0.25 M CH 3 COOH [I] [C]–x–x1+x+x+x+x [E] 0.25–x 1xx 0 + x

x. a) Find the pH of 0.25 M CH 3 COOH [I] [C]–x–x1+x+x+x+x [E] 0.25–x 1xx 0 + x

The concentration of CH3COOH started out as 0.25 a) Find the pH of 0.25 M CH 3 COOH [I] [C]–x–x1+x+x+x+x [E] 0.25–x 1xx

and went down by x, a) Find the pH of 0.25 M CH 3 COOH [I] [C]–x–x1+x+x+x+x [E] 0.25–x 1xx

so its equilibrium concentration will be… a) Find the pH of 0.25 M CH 3 COOH [I] [C]–x–x1+x+x+x+x [E] 0.25–x 1xx

0.25 minus x a) Find the pH of 0.25 M CH 3 COOH [I] [C]–x–x1+x+x+x+x [E] 0.25–x 1xx

The question now is, How do we find x? a) Find the pH of 0.25 M CH 3 COOH [I] [C]–x–x1+x+x+x+x [E] 0.25–x 1xx How do we find x ?

We start solving for x by writing the Ka expression for ethanoic acid a) Find the pH of 0.25 M CH 3 COOH

Using the equilibrium equation, we can see that the Ka expression is equal to the concentration of hydronium times concentration of ethanoate over the concentration of ethanoic acid. a) Find the pH of 0.25 M CH 3 COOH

The equilibrium concentrations of hydronium and ethanoate are both equal to x, a) Find the pH of 0.25 M CH 3 COOH

so for their product in the Ka expression, we can substitute x times x or x squared. a) Find the pH of 0.25 M CH 3 COOH

The concentration of ethanoic acid at equilibrium is x a) Find the pH of 0.25 M CH 3 COOH

So we’ll substitute 0.25 –x in for the concentration of CH3COOH a) Find the pH of 0.25 M CH 3 COOH

The degree of ionization for ethanoic acid is very low. So we make the assumption that x is insignificant compared to This can be written as 0.25 – x is almost equal to 0.25 a) Find the pH of 0.25 M CH 3 COOH Assume 0.25–x  0.25

We can check this assumption later when we determine the % ionization. In general the assumption is valid if the percent ionization is 5% or less. a) Find the pH of 0.25 M CH 3 COOH Assume 0.25–x  0.25 Valid if the Percent Ionization is 5% or Less

Using this assumption will help us avoid having to use a quadratic equation. In Chemistry 12, this assumption is generally used here. When you use it, you Must Always state it. a) Find the pH of 0.25 M CH 3 COOH Assume 0.25–x  0.25 Always STATE this assumption if you’re using it!

so taking out the x on the bottom, we can state that Ka is approximately equal to x 2 over 0.25 a) Find the pH of 0.25 M CH 3 COOH

Rearranging this equation to solve for x 2 gives us x 2 = 0.25 times the Ka a) Find the pH of 0.25 M CH 3 COOH

Taking the square root of both sides gives us x is equal to the square root of 0.25 times the ka. a) Find the pH of 0.25 M CH 3 COOH

At this point, we’ll remind ourselves that x is equal to the equilibrium concentration of hydronium. a) Find the pH of 0.25 M CH 3 COOH

Looking on the acid table, we see that the Ka for ethanoic acid is 1.8 × 10 -5, so we substitute that for Ka in the equation. a) Find the pH of 0.25 M CH 3 COOH

0.25 times 1.8 × 10-5 is equal to 4.5 × 10 -6, so we’ll substitute that in here, under the square root sign. a) Find the pH of 0.25 M CH 3 COOH

The square root of 4.5 x 10-6 is 2.12 x Because we now have a value for the concentration of hydronium, we’ll include the unit Molarity. a) Find the pH of 0.25 M CH 3 COOH

Both the value of Ka and the given concentration have 2 significant figures, which limits our final answer to 2 significant figures. a) Find the pH of 0.25 M CH 3 COOH

We’ve written the concentration of hydronium with 3 significant figures here. We’ll use this and round to 2 significant figures at the end. a) Find the pH of 0.25 M CH 3 COOH

We’re asked for the pH. The pH is the negative log of the hydronium ion concentration, which is the negative log of 2.12 × a) Find the pH of 0.25 M CH 3 COOH

And that comes out to a) Find the pH of 0.25 M CH 3 COOH

But we round this to 2.67 in order to give us 2 decimal places in this pH value, or 2 significant figures in our final answer. a) Find the pH of 0.25 M CH 3 COOH

So now we have the final answer for Part “a”. The pH of 0.25M ethanoic acid is equal to 2.67 a) Find the pH of 0.25 M CH 3 COOH

Part “b” of this question asks us to find the percent ionization of 0.25 M ethanoic acid. b) Find the % ionization of 0.25 M CH 3 COOH

At one point in our calculations, we had determined that the hydronium ion concentration in this solution is 2.12 x molar. b) Find the % ionization of 0.25 M CH 3 COOH

We can use this to help us find the percent ionization b) Find the % ionization of 0.25 M CH 3 COOH

We’ll make a note of it up here. b) Find the % ionization of 0.25 M CH 3 COOH

The formula for percent ionization is the hydronium ion concentration divided by the initial concentration of the acid times 100 percent. b) Find the % ionization of 0.25 M CH 3 COOH

We’ll substitute 2.12 × M in for the concentration of hydronium b) Find the % ionization of 0.25 M CH 3 COOH

And 0.25 M in for the initial concentration of ethanoic acid. b) Find the % ionization of 0.25 M CH 3 COOH

We can cancel the unit Molarity and we multiply by 100% b) Find the % ionization of 0.25 M CH 3 COOH

And we get 0.85% b) Find the % ionization of 0.25 M CH 3 COOH

So now we’ve answered question “b”. The percent ionization of 0.25 M ethanoic acid is 0.85%. Notice it’s quite small, less than 1% of the ethanoic acid molecules have ionized. b) Find the % ionization of 0.25 M CH 3 COOH

remember during the calculations for pH, we assumed that x was insignificant compared to 0.25 M. We said this was valid if the % ionization is 5% or less. a) Find the pH of 0.25 M CH 3 COOH Assume 0.25–x  0.25 Valid if the Percent Ionization is 5% or less

We have just determined that the % ionization is 0.85 %, which is much less than 5%. Assume 0.25–x  0.25 Valid if the Percent Ionization is 5% or less Percent Ionization = 0.85% a) Find the pH of 0.25 M CH 3 COOH

So this verifies that our assumption was definitely valid! Assume 0.25–x  0.25 Valid if the Percent Ionization is 5% or less Percent Ionization = 0.85% a) Find the pH of 0.25 M CH 3 COOH