# The pH of a Mixture of Weak Acids

1. Feb 14, 2009

### 312213

1. The problem statement, all variables and given/known data
2.00M chlorus acid and 2.00M formic acid are mixed. What is the pH?

2. Relevant equations
Ka of chlorus acid is 1.1×10-2
Ka of formic acid is 1.78×10-4

3. The attempt at a solution
For chlorus acid with quadratic equation to find x:
x²+0.011x-0.022
(-0.011$$\pm$$$$\sqrt{}0.000121-4(1)(-0.022)$$)/2
(-0.011$$\pm$$0.297)/2
The positive number of the two choices is 0.143

For formic acid, and the x being small enough to neglect, x is:
x²/2=0.000178
x²=0.000356
x=0.0189

When mixing weak acids, I think I was told that only the acid with the larger Ka value matters, meaning taking the negative log of chlorus acid's H ion gets -log0.143=0.845.

I wasn't sure this was correct so then I tried adding the two H ion concentration from each acid to find their pH. I did 0.143+0.0189=0.162 and negative log that for 0.791.
I'm not sure it is added this way, since I got a bit confused about the idea of adding molarity together, so I divided by 2 first then negative log for 1.09.

Still I'm not sure this is correct so I think I took the wrong steps. How is this correctly done?

2. Feb 14, 2009

### Staff: Mentor

It isn't entirely true - you can do it if the difference between both acids pKas is large enough and if the weaker acid is weak enough. That's the case here and 0.84 is a correct answer.

Unfortunately, it is hard to be more specific.