What is the pH of a Mixture of 2.00M Chlorus Acid and 2.00M Formic Acid?

[312213]

Homework Statement

2.00M chlorus acid and 2.00M formic acid are mixed. What is the pH?

Homework Equations

Ka of chlorus acid is 1.1×10^-2
Ka of formic acid is 1.78×10^-4

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?

 

[Borek]

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.

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.

 

[Blueshift5]

I would like to clarify a few things before providing a response. First, it is important to note that the given statements do not specify the volume of the mixture. This is a crucial factor in determining the pH of a solution. Additionally, the equations and calculations provided do not take into account the dissociation of water, which also affects the pH of a solution.

Assuming that the volume of the mixture is 1 liter, we can proceed with the following steps:

1. Calculate the molar concentrations of chlorus acid and formic acid in the mixture. Since the initial concentrations are given in terms of molarity (M), we can directly use these values. Thus, the final concentration of chlorus acid is 2.00M and the final concentration of formic acid is also 2.00M.

2. Calculate the dissociation constants (Ka) for each acid. This step is not necessary as the given values for Ka are already provided in the statement.

3. Use the Henderson-Hasselbalch equation to calculate the pH of the mixture. The Henderson-Hasselbalch equation is given as pH = pKa + log ([A-]/[HA]), where pKa is the negative logarithm of the dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid. In this case, we have two acids in the mixture, so we need to consider the concentrations of both the conjugate base and the acid for each acid.

For chlorus acid:
pH = 2.95 + log ([ClO2-]/[HClO2]), where [ClO2-] is the concentration of the conjugate base (chlorite ion) and [HClO2] is the concentration of the acid (chlorus acid). Since both acids have the same concentration, the ratio [ClO2-]/[HClO2] will be 1, and thus the log term will be 0. Therefore, the pH of the mixture of chlorus acid is 2.95.

For formic acid:
pH = 3.75 + log ([HCOO-]/[HCOOH]), where [HCOO-] is the concentration of the conjugate base (formate ion) and [HCOOH] is the concentration of the acid (formic acid). Again, since both acids have the