Calculating the pH of a weak acid solution

[Sconesnatcher]

Is there any simple way to calculate the pH of a weak acid in a solution if you already know the dissociation constant of the acid? I've looked it up but can only find long methods that involve factorising quadratic equations. Are these long methods the only way of getting the pH of a weak acid when you know the concentration and dissociation constant of the acid?

[hermanntrude]

Well, there's not always a simple way, but sometimes there's an easier way than solving a quadratic, although you'll need to be able to do both, and both methods involve the ICE table.

 

For instance, if you're asked to find the pH of a 1M solution of a weak acid with a Ka of 1 x 10^-10, you can plug into your ice table like this:

 

Acid + Water <=> Hydronium + base

I 1M - 0 0

C -x - +x +x

E 1-x - x x

 

Notice we ignore the water because it is a pure liquid and has an activity of 1.

 

Now you can plug in your algebraic values for concentrations at equilibrium into your equilibrium constant expression:

 

[math]Ka = 1 \times 10^{-10}[/math] = [math]\frac{x^2}{1-x}[/math]

 

Now you can see that since our Ka is so tiny (one ten billionth), the chances are that subtracting it from 1 is going to give us approximately 1. This means we can make a simplification:

 

[math]Ka = 1 \times 10^{-10}[/math] = [math]\frac{x^2}{1-x}\approx \frac{x^2}{1}[/math]

 

And so in this case we can simply take the square root of the Ka and that will give us a value for x, from which we can calculate the pH.

 

[math] x = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5}[/math]

 

This approximation only works when the value of Ka is very small compared to the initial concentration (in this case 1M). You must ALWAYS plug your value of x BACK into the equilibrium expression to check how close it comes to the real value of Ka:

 

Assumption check:

 

[math]\frac{x^2}{1-x} = \frac{(1 \times 10^{-5})^2}{1-(1 \times 10^{-5})} = 1.00001 \times 10^{-10}[/math]

 

So you see our value for Ka calculated using the assumption is within 0.001% of the actual value, which means the assumption is valid. The rule is that it should be within 5% of the actual value. If it isn't, then you MUST solve the quadratic.

 

It doesn't always mean you have to use the quadratic formula, though. Sometimes, although not usually when working with acids and bases, you'll get a perfect square, ie something like this:

 

Ka = [math]\frac{x^2}{(x-0.1)^2}[/math], in which case you can safely take the square root of both sides.

 

In some cases (perhaps not in your course), you'll find even higher order equations, cubic, quartic or worse. In those cases you will usually either get a perfect cube or something like that, or you'll get a simplification made obvious to you, although there are methods for solving cubics and higher using successive approximations (feel free to look them up if you think you might need them, but it's rare).

 

Bear in mind, though, that most courses DO require you to be able to solve quadratics using the quadratic formula, so get used to it. practice practice practice :0)

 

PS sorry about the ICE table... it didn't work out as planned. Anyone know how to insert a table?

[Sconesnatcher]

Ah right I get it now. So the H+ concentration is always equal to its conjugate base concentration?

 

So to get the question done fast on the test I can just use the formula squareroot(Ka x INITIAL CONCENTRATION)?

 

For example lets say theres 0.2M of an acid with a Ka of 1.8 x 10^-4. If I just multiply 1.8 x 10^-4 by 0.2 I get the equation x^2 = 3.6 x 10^-5. If I get the square root of that then I have x = 0.006. Is that right?

[hermanntrude]

Ah right I get it now. So the H+ concentration is always equal to its conjugate base concentration?

 

So to get the question done fast on the test I can just use the formula squareroot(Ka x INITIAL CONCENTRATION)?

 

For example lets say theres 0.2M of an acid with a Ka of 1.8 x 10^-4. If I just multiply 1.8 x 10^-4 by 0.2 I get the equation x^2 = 3.6 x 10^-5. If I get the square root of that then I have x = 0.006. Is that right?

 

NO. it doesn't always work out the same. You have to follow these steps:

 

-draw out the equation

-draw out the ICE table

-write out the Ka expression in terms of x

-check to see if the assumption is likely to work, if it is, make the assumption, get x and check to see if the assumption gave an error of less than 5%

-if the assumption doesnt work, check to see if it's a perfect square, and if not, use the quadratic formula.

 

There is no quick way around this, but if you learn the method it can be quite easy with practice.