# Homework Help: Titration with weak acid and strong base

1. Dec 2, 2007

### stardusto12

1. The problem statement, all variables and given/known data
We did a lab in class about weak acid/strong base titration, but I'm having a bit of trouble. We we only given the molarities of an unknown weak acid and a strong base (NaOH) and the pH of six different mixtures containing different mL of the two, which is below:

Test tube #1: 9mL of weak acid and 1mL of NaOH pH=3.04
Test tube #2: 8mL of weak acid and 2mL of NaOH pH=3.07
Test tube #3: 7.5mL of weak acid and 2.5mL of NaOH pH=3.40
Test tube #4: 7mL of weak acid and 3mL of NaOH pH=3.56
Test tube #5: 6mL of weak acid and 4mL of NaOH pH=3.80
Test tube #6: 5.5mL of weak acid and 4.5mL of NaOH pH=4.21

and the molarity of the solutions were .1176 for the unknown acid and .1059 for NaOH.

2. Relevant equations
I was able to calculate the intial moles of the acid and base but were I'm having difficulty is figuring out the moles at equilibrium. (of A- and HA in HA + H2O = H3O +A-).

3. The attempt at a solution
I'm very confused and was only able to get the intial moles of the acid and base. I know I'm supposed to use the Henderson-Hasselbaclh equation, but I don't know how to get log of (HA/A-) since I don't have HA or A-.

Thanks!

2. Dec 2, 2007

### symbolipoint

The formalities can be obtained directly from how each solution is prepared. What is the question which you are trying to answer? Are you trying to find a value for the equilibrium constant for the weak acid?

3. Dec 2, 2007

### stardusto12

Yes, I am trying to find the equilibrium constants, but of both. I am trying to find the equilibrium concentrations of the both the acid and base. Afterwards, I can calculate the rest on my own. I just don't know how to derive the equilibrium concentrations from the information given.

Last edited: Dec 2, 2007
4. Dec 2, 2007

### symbolipoint

K = (H)(Fs + H)/(Fa - H), The equilibrium constant expression for a weak acid.

H = the hydronium ion concentration as molarity
Fs = formality of the salt of the weak acid
Fa = formality of the weak acid

Examine that formula carefully to see that the molarity of the salt is Fs + H and the
molarity of the weak acid is Fa - H. The formula should be just fine for you because you
are dealing with only acidic solutions of the weak acid. I would not call your data a set
from an actual titration; the values seem to look wrong for that. You should still be able to
use each one of them to find any unknown value for the K formula.

5. Dec 2, 2007

### stardusto12

thank you! Oh, what do you mean by formality? and is it the ph that's wrong? I believe my partner didn't clean it off well with the distilled water....

6. Dec 2, 2007

### symbolipoint

Formality means the concentration of a substance as formula units per liter based on the way the solution was prepared, with no regard for reactions of the substance - maybe not the best way to describe the meaning of Formality, but to be certain, check the meaning of this in an analytical chemistry textbook.

If you mix 1 mole of acid HA with 0.25 mole NaOH, then you still began with 1 mole of HA. You now have 0.75 moles unneutralized HA and 0.25 moles of NaA.

If you mix 100 ml of 1 M HA with 25 ml of 1 M NaOH, then you have still
0.100/125 Formal HA (you can ignore the reaction, but the MOLARITY will be different).

7. Dec 2, 2007

Thanks!

8. Dec 2, 2007

### symbolipoint

Actually, I stated the latter part wrongly.
The formality would be (0.100 liter)*(1 mole/liter)/(0.125 liter) for the acid HA.