[SweetAsC12H22O11]

A 50mL solution containing 0.10M 1-Naphtoic acid (pKa=3.70) and 0.15M arsenic acid (pKa1=2.24, pKa2=6.96) is titrated with 0.20M KOH.

a) How many end points will be observed?
b) What is the pH of the solution before adding any KOH?
c) Calculate the volume of KOH required to reach the first equivalence point.
d) Calculate the pH after addition of 100mL of KOH.

My "attempts"

a)
Since pKa of 1-Naphtoic acid is within 3 units of pKa1 of arsenic acid, while pKa2 of      arsenic acid  is more than 3 units greater, there are 2 end points.

b)
How would I determine [H⁺] in order to calculate pH?

   Normally, at 0mL of titrant added, [H⁺] is calculated using,

   Ka = ([A^-][H⁺])/[HA] = x²/(F-x)

For concentration F in this equation, would I use the sum of concentrations of each      acid of 0.25M? I started by calculating the initial number of moles of each acid (since no titrant was added yet), adding them together, and dividing the sum by 0.05L, which gave me the sum of the concentrations...0.25M. Also, what value would I use for Ka since both pKa's represent 1 end point?

c)
If I use the sum of concentrations (which I'm really not sure makes sense), then:

     (0.25M)(0.05L) = (0.20M)V_e ---> V_e = 0.0625L = 62.5mL

d)
The number of moles of KOH added at 100mL is 0.02 moles

The initial moles of weak acid = 0.005 moles + 0.0075 moles = 0.0125 moles

And...Well... I just don't understand mixed titrations and have nothing about them in my textbook... Any help would be very much appreciated....
   

[Borek]

Since pKa of 1-Naphtoic acid is within 3 units of pKa1 of arsenic acid, while pKa2 of      arsenic acid  is more than 3 units greater, there are 2 end points.

More or less correct, although these predictions are rarely accurate. Can be the first end point is barely visible.

How would I determine [H⁺] in order to calculate pH?

Tricky. I can't think of an easy and reasonably accurate approximation that won't yield 3rd degree polynomial.

c)
If I use the sum of concentrations (which I'm really not sure makes sense), then:

     (0.25M)(0.05L) = (0.20M)V_e ---> V_e = 0.0625L = 62.5mL

Trivial stoichiometry. Think which protons are neutralized up to the first endpoint.

d)
The number of moles of KOH added at 100mL is 0.02 moles

The initial moles of weak acid = 0.005 moles + 0.0075 moles = 0.0125 moles

That means substantial excess of KOH, so you probably can ignore presence of other (much weaker) bases in the solution.

[SweetAsC12H22O11]

Okay thank you, so I think I understand a little more, tried again:

b)
An acid with a lower pKa dissociates at a lower pH, so low pKa=low pH = more acidic, right? So the arsenic acid (pKa1) will be the first deprotonated. That must mean that the pH at 0mL of titrant added can be calculated using Ka1 of arsenic acid. If this is correct, then:

Ka1 = ([A^-][H⁺])/[HA] = x²/(F-x)

(5.7544 x 10^-3) = x²/(0.15 - x)

Use quadratic to find x = 0.02664M = [H⁺], therefore pH=1.574


c)
I think both 1-Naphtoic acid (NA) and arsenic acid (AA) are deprotonated, NA being first. At V_e, the moles of the neutralized acids = the moles of KOH used. So since both are neutralized up to the first end point:
     
     total moles = 0.005 + 0.0075 = 0.0125 moles = moles KOH

     0.0125 moles / 0.2(moles/L) = 0.0625L = 62.5mL


d)
to come...

[SweetAsC12H22O11]

d)

If the first EP is at 62.5mL, then when KOH added is 100mL, we are in between 2 EP's. All the NA should be titrated now and only AA remaining (second deprotonation).

In between the 2 EP's, the number of moles of AA = 0.005

Between the 1st EP and 100mL of KOH : 100 - 62.5 = 37.5mL KOH = 0.0075 moles

Why is OH in excess in between to EP's? Because we are closer to the second EP (125mL) than the first?

Since OH^- is in excess, the moles remaining of KOH is 0.0075-0.005 = 0.0025moles

Total volume = initial 20mL of weak acid + 100mL KOH = 120mL = 0.120L

So [OH^-] = 0.02083M   which mean pOH = 1.681   and pH = 12.319

This seems high... but then again there is an excess of OH...

[Borek]

An acid with a lower pKa dissociates at a lower pH, so low pKa=low pH = more acidic, right? So the arsenic acid (pKa1) will be the first deprotonated. That must mean that the pH at 0mL of titrant added can be calculated using Ka1 of arsenic acid. If this is correct, then:

Ka1 = ([A^-][H⁺])/[HA] = x²/(F-x)

(5.7544 x 10^-3) = x²/(0.15 - x)

Use quadratic to find x = 0.02664M = [H⁺], therefore pH=1.574

This is a very good first approximation. Generally speaking pH will be slightly lower, but now that we have a solid number to talk about, pH is two over units from pKa of naphthoic acid, so the naphthoic acid is dissociated in less than 1% - so we can safely ignore its effect.

c)
I think both 1-Naphtoic acid (NA) and arsenic acid (AA) are deprotonated

No doubt about it, question is - how many protons of AA are neutralized?

Actually now I realized your earlier analysis completely ignored the fact AA is multiprotic, so your stoichiometry is off (especially for point d). You have 0.1+3*0.15 protons to be neutralized per liter of the solution, not 0.1+0.15, so there is no excess of KOH. Sorry for missing it.

[SweetAsC12H22O11]

Okay, I'm not sure I understand. So because of the stoichiometry, the number of moles of AA should be multiplies by 3?

c)
moles NA = 0.1M x 0.05L = 0.005moles
moles AA = 3(0.15 x 0.05L) = 0.0225moles

total = 0.0275 moles = moles KOH / 0.2M = 0.1375L = 137.5mL = V_e?

that seems very high... I must be missing something

d)

If this is the case for d, then okay, there is no excess of KOH.

moles of AA (HA) between 2 EP's = 0.0225
mole of KOH added at 100mL = 0.0075

So moles of HA remaining is 0.015 and moles of A^- is 0.0075.

Can we use HH equation to find the pH?

pH = pKa1 + [A^-]/[HA] = 2.24 + (0.0075/0.0225) =2.573?

I feel I'm getting more and more confused...

[SweetAsC12H22O11]

I should metion that for this question, the prof. said to omit the third pKa of arsenic acid and say it is diprotic.

Does this mean we have 0.1 + 2*0.15 protons to neutralize?

[SweetAsC12H22O11]

If what I wrote in my most recent post is right, does this mean then that moles of AA should be multiplied by 2 since it is "diprotic"?

in which case...

c)
moles NA = 0.005, moles AA = 0.015, total = 0.02 = moles KOH

multiply by conc. of KOH (0.2M) = 0.1L = 100mL = V_e

d)
If V_e = 100mL, then we are at V_e for this questions, which means:

pH = 0.5 * (pKa1 + pKa2) = 0.5 * (2.24 + 6.96) = 4.6

However, this seems to ignore the amount of NA in the mixture...

[Borek]

Think about it this way: when you add base, first, the strongest acid is neutralized - that means first proton of the arsenic acid. Then, naphtoic acid*. At this moment you come to the first endpoint - so you should be able to calculate amount of KOH needed from the stoichiometry. How many moles of protons from the arsenic acid? How many moles of protons from the naphtoic acid?

After that the second proton of the arsenic acid becomes available and becomes neutralized. How far? Depends on the amount of KOH added. Some of the KOH was already used to neutralize arsenic acid first proton and naphtoic acid. How much more was added? How far did the neutralization of the second proton of the arsenic acid went? (Be ready for a surprise.)

*actually it is not that simple, as they are a little bit too close to each other to be treated completely separately, but it is a good approximation that should help you understand what is happening.