Go to a short intro to what K_{a} is

Go to Solving K_{a} Problems: Part Two

Go to Solving K_{a} Problems: Part Three

Return to a listing of many types of acid base problems and their solutions

Important note: all constants refered to: K_{c}, K_{w}, K_{a}, and K_{b} are temperature-dependant. All discussions are assumed to be at 25 °C, i.e. standard temperature.

The typical weak acid problem to solve in high school classes looks like this:

What is the pH of a 0.100 M solution of acetic acid? KSome facts of importance:_{a}= 1.77 x 10¯^{5}

1) you know this is a weak acid for two reasons:a) you memorized a short list of strong acids. (You did, didn't you?) Everything else is weak.2) The solution technique explained below applies to almost all weak acids. The only things to change are the concentration and the K

b) The K_{a}value of small. strong acids have very large K_{a}values, as in 10^{5}or 10^{7}._{a}, if doing another acid.

You've seen these equations:

HAc + H_{2}O ⇌ H_{3}O^{+}+ Ac¯

K_{a}= ( [H_{3}O^{+}] [Ac¯] ) / [HAc] = 1.77 x 10¯^{5}

The key quantity we want is the [H_{3}O^{+}]. Once we have that, then the pH is easy to calculate. Since we do not know the value, let's do this:

[H_{3}O^{+}] = x

I hope that, right away, you can see this:

[Ac¯] = x

This is because of the one-to-on molar ratio between [H_{3}O^{+}] and [Ac¯] that is created as HAc molecules dissociate.

So now, we have all but one value in our equation:

1.77 x 10¯^{5}= {(x) (x)} / [HAc]

All we have to do is figure out [HAc] and we can calculate an answer to 'x.'

In this problem, the [HAc] started at 0.100 M and went down as HAc molecules dissociated. In fact, due to the one-to-one ratio, it went down by 'x' amount and wound up at an ending value of 0.100 - x.

So, here's the final set-up:

1.77 x 10¯^{5}= {(x) (x)} / (0.100-x)

Now, that is a quadratic equation and can easily be solved with the quadratic formula. However, there is a trick we can use to make our calculation easier.

You don't know it, but K_{a} values are very difficult to figure out.There's a whole bunch of variables that are difficult to control. The end result is that K_{a} are approximate and most are in error about ± 5%.

There is a brief discussion of the 5% Rule here.

So that means, if we stay within 5% of the answer using the quadratic, we can use approximate techniques to get an answer. The major approximation occurs with '0.100 - x.' Since x is rather small, it will not change the value of 0.100 by much, so we can say:

0.100 - x approximately equals 0.100

We now write a new equation:

1.77 x 10¯^{5}= {(x) (x)} / 0.100

which is very easy to solve. We move the 0.100 to the other side to get:

x^{2}= 1.77 x 10¯^{6}

Taking the square root (of both sides!!), we get:

x = 1.33 x 10¯^{3}M

Take note of two things:

1) The K_{a}value is unitless, but x is a molarity.

2) Square root both sides. I have had students square root the x^{2}, but not the other side. Weird, but true.

We finish by taking the pH to get a final answer of 2.876.

The final comment has to do with checking for 5%. The formula is:

( [H_{3}O^{+}] / [HAc] ) x 100 < 5%

In our case, we had 1.33%, which is acceptable.

Go to a short intro to what K_{a} is

Go to Solving K_{a} Problems: Part Two

Go to Solving K_{a} Problems: Part Three

Return to a listing of many types of acid base problems and their solutions