What is the relationship between titrations and starting pH?

[brake4country]

Homework Statement

What was the concentration of acid S before the titration?

Homework Equations

M_aV_a=M_bV_b

The Attempt at a Solution

So for this problem, the solution manual suggests that we use the formula above to solve for the pH of acid S before the titration. But why do we use the concentration of the base before we even add it? Wouldn't it be easier to locate the pH on the graph (which is approximately 3.8) and take 10^-3.8 = 1.6 x 10^-4?

The correct answer is B. Thanks in advance.

 

[BvU]

pH doesn't help for the concentration if the acid isn't completely dissociated into H+ and Ac-.
What concentration would you find for acid S from the pH ?
What would you find for acid T ? pH and concentration ?

 

[Borek]

We don't care about initial pH at all, just about the stoichiometry of the reaction. Do you know how the titration works? Why is part of the curve so steep?

 

[brake4country]

Yes, I understand how it works. The steepest part of the graph is where the equivalence point i and where the pH rises the fastest. Contrarily, the half equivalence point (the horizontal portion) is where the pH changes the least with increased base. But how are we to determine the stoichiometry of the reaction when we know nothing about the acid?

 

[brake4country]

pH doesn't help for the concentration if the acid isn't completely dissociated into H+ and Ac-.
What concentration would you find for acid S from the pH ?
What would you find for acid T ? pH and concentration ?

That is where I am confused. The directions say that we start with 50 mL of acid S and the titrant is 0.1 M NaOH. With M_A as the unknown, I came up with:
M_AV_A=M_BV_B
(0.050L)(M_A) = (0.1)(V_B)

But why incorporate any base when the question asks "what is the concentration of acid S before the titration"?

 

[BvU]

Don't want to spoil your exercise. Try to move on to 814 and perhaps you'll see. (and note Borek's post #3)

 

[brake4country]

I think I see what is going on here. The concentration of S is unknown and we have to use the volume of the NaOH added at the equivalence point to determine that.
So our M_A= what we are solving for
V_A=(50 mL)
M_B=0.1 M
V_B=50 mL)
M_AV_A=M_BV_B
M_A=0.1 M

The question was a bit misleading but I think I get it now.

 

[brake4country]

Don't want to spoil your exercise. Try to move on to 814 and perhaps you'll see. (and note Borek's post #3)

I understand 814. We can find the Ka from the pKa at the half eq. point. So, if pH = 4 at half eq. point, then ka = 10^-pH = 1x10^-4.

 

[BvU]

Why at the half-eq point ? They ask for the K_a of the acid ...

 

[brake4country]

We don't care about initial pH at all, just about the stoichiometry of the reaction. Do you know how the titration works? Why is part of the curve so steep?

Right. Because that is what is unknown. Haven't done too many titrations before but I think I get it. We can only calculate the unknown starting concentration of S when we reach the equivalence point.

 

[brake4country]

Why at the half-eq point ? They ask for the K_a of the acid ...

Right. So I used Henderson Hasselbach equation and converted pKa to Ka. Have to be quick on timed exams. Is there another way to calculate the Ka without using HHB?

 

[Borek]

Reading pKa from a half-equivalence seems to me the best approach here.

 

[BvU]

OK, sorry. Out of my league. Found http://chemistry.oregonstate.edu/courses/ch421/Previous%20pages/Course%20Docs%20Fall%202006/Henderson.pdf interesting !