K_a : Solving Weak Acid Problems: Part One

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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? K_a = 1.77 x 10¯⁵

Some facts of importance:

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.
b) The K_a value of small. strong acids have very large K_a values, as in 10⁵ or 10⁷.

2) The solution technique explained below applies to almost all weak acids. The only things to change are the concentration and the K_a, if doing another acid.

You've seen these equations:

HAc + H₂O <==> H₃O⁺ + Ac¯
K_a = ( [H₃O⁺] [Ac¯] ) / [HAc] = 1.77 x 10¯⁵

The key quantity we want is the [H₃O⁺]. Once we have that, then the pH is easy to calculate. Since we do not know the value, let's do this:

[H₃O⁺] = x

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

[Ac¯] = x

This is because of the one-to-on molar ratio between [H₃O⁺] 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%.

So that means, if we stay within %5, 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:

x² = 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 x², 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:

( [H₃O⁺] / [HAc] ) x 100 < 5%

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

Go to Solving K_a Problems: Part Two

Go to Solving K_a Problems: Part Three

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