[Roddy]

I need help with two different problems.  I think I may have solved them, but need some confirmation that I did them correctly:

Problem 1:
Calculate the molarity of NaOH solution which requires 44.62 mL to titrate 1.730 grams of thepure oxalic acid, H₂C₂O₄.2H₂O to the phenolphthalein end point.
My attempt:
On this problem I converted oxalic acid to moles= 3.07x10^-1 M H₂C₂O₄.2H₂O
In calculating the end point it is my understanding that you need [I-]/[HI]=1/10 and setting this to [NaOH]/(3.07x10^-1) to solve for [NaOH].  I did this and got 3.7x10^-2 M NaOH as the final answer. Is this correct?
Problem 2:
A weak diprotic acid has a cesium salt, CsHA which is very slighlty soluble.  You titrate 100.0 mL of the saturated solution with 0.0499 M NaOH and reach an endpoint at 17.30 mL.  One of your data points is at 8.30 mL and the pH recorded for it is 4.88.  What is the solubility product, K_sp, and the K_a2 for this compound CsHA?
My attempt:
Again, I tried solving for it and used M_aV_a=M_bV_b
V_a= 100.0 mL
M_a=0.0499
V_b=8.30 mL
and my answer yielded 4.14x10^-3 M CsHA

I then set up the equation as CsHA + H₂O H₃O + CsA^-
[H]= 10^-4.88 = 1.32x10^-5
Ksp = [H₃0⁺][CsA] = (1.32x10^-5)2=1.74x10^-10
For Ka I used H&H equation where 4.88=pK_a + log(8.30/(17.3-8.3))
pK_a=4.92 and K_a=1.2x10^-5

Does this look like I set it up correct?
Thanks

[Borek]

In calculating the end point it is my understanding that you need [I-]/[HI]=1/10 and setting this to [NaOH]/(3.07x10^-1) to solve for [NaOH].

Please eleaborate - I think I know what you are aiming at, but I don't want to guess.

I did this and got 3.7x10^-2 M NaOH as the final answer. Is this correct?

No, you are not in a correct ballpark.

V_b=8.30 mL
and my answer yielded 4.14x10^-3 M CsHA

No, not 8.30 for the titration stoichiometry.

[Roddy]

Please eleaborate - I think I know what you are aiming at, but I don't want to guess.

From what I understand, for most indicators, one tenth of the initial form must be converted to the other form before a new color is apparent (endpoint).  This is usually represented as [In^-]/[HIn]=1/10


And how about Problem 2.  Does anyone have any suggestions on this one??

[Borek]

OK, that's what I guessed. Note: you are probably not asked to cacluate volume that precisely (although it can be done, see http://www.titrations.info/acid-base-titration-end-point-detection and many other pages on that site). My bet is that they just mean oxalic acid was titrated till BOTH protons were neutralized.

To be honest I have not analyzed your answers after spotting first mistakes. Correct problems pointed out.

[Roddy]

Problem 1 (second attempt):

The other way I solved the first problem was as follows:

1.730g Oxalic/126.1g= 1.372x10^-2 mol Oxalic Acid
H₂C₂O*2H₂O + 2NaOH Na₂C₂O₄ + 4H₂O
Ratio is 1:2 therefore moles of NaOH= 1.372x10^-2 x 2 = 0.0274 mol
M=0.02744/0.04462L= 0.615 M NaOH

I thought this answer is the equivalence point and not the endpoint.  I understand that in the perfect world they would be the same, but the fact that the problem asked for end point and not equivalence point I was thrown off.

[Borek]

Difference between both equivalence and end point would be in the range of 0.01 mL.

[Roddy]

Problem 2: (from first post above)

No, not 8.30 for the titration stoichiometry.

I am looking at this several different ways, and can't seem to figure it out.
                                                _ksp
I wrote the equation as CsHA+H₂O  H₃O + CsA^-
I am trying to solve this by using an ICE table with [H⁺]_equilib=10^-4.88 which would also be the same as [A^-]
Ultimately all of this is to get the [CsHA] to determine Ka. 

I am not sure if I am even on the right track with this and if so, where do I go from here?

[Borek]

First of all, to calculate concentration of dissolved CsHA you need to correctly calculate results of titration. What was the volume of NaOH required to fully neutralize HA^-?

[Roddy]

Endpoint volume would be 117.30 mL

[Borek]

That's total, not NaOH.

[Roddy]

OH, I see NaOH volume needed to reach endpoint is 17.30 mL.
If I multiply that by 0.0499M NaOH and devide by total volume of 117.3 mL that should give me the [NaOH]=7.36x10^-3

[Borek]

Geez... you don't need NaOH concentration, you need CsHA. This is simple, obvious, easy titration calculation, not more difficult than HCl + NaOH at this stage.

[Roddy]

I understand that. As I stated in a previous post:

I am trying to solve this by using an ICE table with [H⁺]_equilib=10^-4.88 which would also be the same as [A^-]
Ultimately all of this is to get the [CsHA] to determine Ka.

I know I need [CsHA], but don't know how to get it... hence the reason I am even here posting this. I am not here to quiz you and see if you can get it.  I don't know how to get it and am here asking how!

[Borek]

You don't need ICE table now.

You have saturated solution of CsHA - that means solution containing Cs⁺ and HA^-.

You are titrating 100 mL of solution of a weak acid HA^-. You used 17.30 mL of 0.0499 M NaOH. What is concentration of HA^-?