Titration help (again)

Factual Questions
1

This is not a required assignment.

Consider the titration of 25.0 mL of 0.125 M HCL with 0.100 M KOH. Calculate the pH after the addition of 5 mL base, 20 mL base, and 65 mL base. Let’s find the number of moles of HCL:

25 mL x 1/1000 L x 0.125 mol/L = 0.003125 mol HCl

OK, we start out with 0.003125 mol HCl, and we add 0.0005 mol (5 mL x 1/1000 L x 0.100 mol/L), 0.002 mol, and 0.065 mol KOH. After the first addition, we have 0.002625 mol HCL remaining. We divide that by (.025 L + .005 L) solution to get K[sub]a[/sub] = [H[sup]+[/sup]] = 0.0875, so pH = -log(0.0875) = 1.1. Similarly, the pH after the second addition of KOH is 1.6.

Now here’s where I have the problem. We start with 0.003125 mol HCl and we add .0065 mol KOH. So we ‘have’ negative .003375 mol HCl remaining. Can’t have a negative concentration, so how much KOH do I have? 0.0065 mol KOH reacting with 0.003125 mol HCl leaves 0.003375 mol KOH in 0.09 L of solution for a molarity of .0375 M. The negative log of that is 1.4 pH. That can’t be right, because the solution should be basic.

Where am I going wrong?

2

Too late to edit.

K[sub]a[/sub] x K[sub]b[/sub] = K[sub]w[/sub]. So if K[sub]a[/sub] = 1.0x10[sup]-14[/sup] / .0375 = 2.667x10[sup]-13[/sup], then pH = -log(2.667x10[sup]-13[/sup]) = 12.6, right?

3

When you took the negative log of 37.5 mM, you ended up with pOH. In this rough approximation, pH + pOH = 14.

In real life, you have some extra salts that will affect chemical activities. Although at these concentrations it probably doesn’t matter much.

4

Which gives me 14 - 1.4 = 12.6, which is what I got using the relationship between the K values. Thanks! :slight_smile:

For my next trick, titrating 40.0 mL of 0.250 M HF with 0.200 M NaOH and calculating the pH. K[sub]a[/sub] = 3.5x10[sup]-4[/sup] and K[sub]b[/sub] = 2.9x10[sup]-11[/sup]. I know HF is a weak acid, and that 40 mL of 0.250 M HF is 0.01 mol HF. I think I need to use the equation:

K[sub]a[/sub] = [A[sup]-[/sup]] / [HA]

If I add 8 mL = .008 L 0.200 M NaOH, then I have 0.0016 mol NaOH. If I have 0.0016 mol NaOH, then I have 0.01 - 0.0016 = 0.0084 mol HF remaining and I’ve made 0.0016 mol F[sup]-[/sup] ions in 0.048 L of solution.

[HA] = [HF] = 0.0084 mol / 0.048 L = 0.175 M
[A[sup]-[/sup]] = [F[sup]-[/sup]] = .0016 mol / 0.048 L = 0.3333 M
K[sub]a[/sub] = 3.5x10[sup]-4[/sup], so pK[sub]a[/sub] = -logK[sub]a[/sub] = 3.46.

pH = pKa + log([F[sup]-[/sup]]/[HF]) = 3.46 + log(0.333/0.175) = 3.7

Something doesn’t look right… :confused:

5

Some of your equations are off. Take a peek at this website:

Especially re: the acid dissociation constant, which is typically not unitless. For HFsub[/sub] it’s 3.5x10[sup]-4[/sup] mol L[sup]-1[/sup]. In your equation the units cancel out.
HF, of course, is weird because it forms chains and homoassociates, but again, we’re just approximating here.

6

Some of my equations? Which ones? I know ‘something’ doesn’t look right, but I can’t see what.

7

K[sub]a[/sub] = [H[sub]3[/sub]O[sup]+[/sup]][A[sup]-[/sup]] / [HA]

8

I have 0.0084 mol HF in 0.048 L solution, so I have 0.175 M HF. I have 0.0016 mole F[sup]-[/sup] in 0.048 L, so I [A[sup]-[/sup]] is 0.333. We’re given K[/sub]a[/sub] as 3.5x10[sup]-4[/sup] So…

3.5x10[sup]-4[/sup] = [H[sub]3[/sub]O[sup]+[/sup]] x 0.333 / 0.175, and [H[sub]3[/sub]O[sup]+[/sup]] = 3.5x10[sup]-4[/sup] x 0.175 / 0.333 = 1.84x10[sup]-4[/sup]. That gives me a pH of 3.7.

Is that right?

9

0.0016/0.048 = 0.03333…

Your division is wrong.

10

Thanks. I’ll recalculate mañana.