[Anthasci]

Greetings everyone,

I am currently struggling trying to understand buffer systems and more generally speaking, neutralising titrations of all varieties. Specifically, I came across this problem;

1.) Determine the pH of buffer solutions obtained  by mixing 50,00 mL of 0,200M NaH₂PO₄ with;
a) 50,00 mL of 0,120M HCl (1,97)
b) 50,00 mL of 1,120M NaOH (7,37)
K₁=7,11*10^-3
K₂=6,34 *10^-8
K₃=4,2*10^-13

for a), my idea was to write the following equation:
NaH₂PO₄ + HCl   H₃PO₄ + NaCl
which to my knowledge, can be regarded as a one-way?
Next, I wrote:
H₃PO₄ + H₂O   H₂PO₄^- + H₃O⁺

So, NaH₂PO₄ and HCl react at a molar ratio of 1:1, which is 0,006 mol (some sodium dihydrogenphosphate is left out), resulting in a 0,06M solution of H₃PO₄ (total volume is now 100 mL)

Using the K₁, I figured out that in the second equation, [H₃O⁺] equals 0,017403, which in turn, results in the pH of 1,76. Did I go wrong somewhere or are the solutions wrong (written above)? Also, I'm guessing b) has pretty much the same idea? I got 6,68 for b) but I think this is wrong, since I have a strong feeling the system would have a basic pH.

Thanks!

[Borek]

It is not clear to me how you got your answers. You are right assuming H₂PO₄^- reacts with added strong acid and strong base and trying to calculate concentrations of HPO₄^2- and H₃PO₄ (but look below), but it is not clear to me what you do later. Common approach to the problem would be to put these concentrations into the dissociation constant definition - Ka is known, concentrations of acid and conjugate base are known, the only thing unknown is [H⁺]. Dissociation constant definition can be rearranged to the so called Henderson-Hasselbalch equation, which is very convenient for this type of calculations.

Unfortunately, this approach won't work for a, and answer given for a is wrong (pH is not 1.97, more like 2.11). H₃PO₄ is simply a little bit too strong acid for the assumption that H₂PO₄^- gets quantitatively protonated by HCl, and finding the answer requires more complicated calculations (not THAT complicated - ICE table will do).

Answer given for b is correct, can't comment on your result as I have no idea how you got it.


Edit: I think I know what you did - you calculated pH using concentration of protonated H₃PO₄ only. You can't - solution already contained conjugate base, when you calculate pH starting from known concentration of an acid you assume initial concentration of the conjugate base to be zero.

[AWK]

b) 50,00 mL of 1,120M NaOH (7,37)

? should be 0.120 instead 1.120 ?
Think about stoichiometry of neutralization in both cases. (both suggested answers are correct)

[Borek]

(both suggested answers are correct)

1.97 is incorrect. You can get it applying blindly HH equation, but it doesn't make it a correct answer.

[Anthasci]

Borek - I don't think I understand. For a), you say that the concentrations of acid and conjugate base are known, but which numbers are they, here? Was I correct in saying that the H₃PO₄ is 0,06M? So, since 0,004 moles of NaH₂PO₄ are left over, does this mean it's concentration is now

[itex] \frac {0,004 mol}{0,1 L} [/itex]= 0,04M (A-)?

Then I'd go [H₃O⁺]=K_a1[itex] \frac {0,06}{0,04} [/itex]

I understand this is a simplyfied solution, and probably what the prof. had in mind for us. Alternatively, what would be the correct solution?

Then I went over to b) and got lost - I see that from the molar ratio, some NaOH would be left unreacted - does this mean it dissolves in water and gives me a certain concentration of [OH^-]? Calculating the [H₃O⁺] from the dissociation of the acid, I can then find out how much [OH^-] is left, since the reaction that takes place is:

OH^- + H₃O⁺   2H₂O

[Borek]

Then I'd go [H₃O⁺]=K_a1[itex] \frac {0,06}{0,04} [/itex]

What you did to calculate concentrations of H₃PO₄ and H₂PO₄^- was OK and would work for a slightly weaker acid with pKa of about 3 or higher.

I understand this is a simplyfied solution, and probably what the prof. had in mind for us. Alternatively, what would be the correct solution?

Note that assuming all phosphoric acid is in the form H₃PO₄ you are assuming there is no H⁺ in the solution, as it was all used to protonate the acid. But you just calculated that pH is 1.97, so you know phosphoric acid concentration is not 0.06 - which in turn means pH is not 1.97. For weaker acids this difference in insignificant, here is gives a relatively large error.

It can be relatively easy dealt with using ICE method.

     H₃PO₄     H⁺    +    H₂PO₄^-

I     0.06            0            0.04
C      -x             +x           +x
E     0.06-x         x            0.04+x

[tex]K_{a1} = \frac{x(0.04+x)} {0.06-x}[/tex]

Solve for x - that's the H⁺ concentration, and it yields the correct pH.

Then I went over to b) and got lost - I see that from the molar ratio, some NaOH would be left unreacted - does this mean it dissolves in water and gives me a certain concentration of [OH^-]?

I think you have a typo in your data and it should be 0.120, not 1.120 (otherwise you will not get 7.37 as the correct answer). But if 1.120 is correct, it means it is excess NaON that is responsible for the pH (especially when the excess is large).

[Anthasci]

Thanks for the clarification!

Yeah, I suppose it makes more sense that c_NaOH=0,120M

Though I am confused ... do I still keep working with the K_a in the b) problem, or K_b, since the pH > 7? I assume all of NaOH is used in the reaction

NaH₂PO₄ + NaOH Na₂HPO₄ + H₂O

leaving me with a 0,06M solution of Na₂HPO₄ and a 0,04M NaH₂PO₄, since 0,004 moles out of 0,01 moles don't react and the total volume is 100 mL.

Using the same approach as before, based on the [H₃O⁺], I get a pH of a little over 12.

[Borek]

Yes, you still work with pKa.

Show how you got 12, concentrations are OK but 12 is not.

[Anthasci]

Ok, so this is the first equation:

NaH₂PO₄ + NaOH Na₂HPO₄ + H₂O

Then:

HPO₄^2- + H₂O   PO₄^3- + H₃O⁺

The K_a3 is 4,2*10^-13

c_HPO4^2- = 0,06M
c_PO4^3- = 0,04M

Actually, I'm not really sure whether to use K_a2 or K_a3 here

or does the leftover NaH₂PO₄ also dissolve in water or something like that?

EDIT: Solved it.

H₂PO₄^-   H⁺ + HPO₄^2-

knowing 0,04M H₂PO₄^- and 0,06M HPO₄^2-, I just solved it using ICE. Nice. Thank you very much for your aid.

[Borek]

K_a2, you have H₂PO₄^- and HPO₄^2- in the solution, this is the second dissociation equilibrium.

[tex]pH = pK_{a2} + log\frac{[HPO_4^{2-}]}{[H_2PO_4^-]}= pK_{a2} + log\frac{0.06}{0.04} = 7.37[/tex]

[Anthasci]

Ok, so technically speaking, this following question is related to general equilibrium and solubility product, but I figured I'd just throw it in my existing topic.

5. Find the solubility of ZnC₂O₄ (g/L) at pH = 3
ZnC₂O₄   Zn²⁺ + C₂O₄^2-
K_sp=7,5*10^-9
K_b1=1,8*10^-10
K_b2=1,8*10^-13

If I'm correct, s = [Zn²⁺] + [C₂O₄^2-]

Obviously the pH is acidic and I'm confused why I'm given the K_b. [itex] \frac {Kw}{Kb_1} = Ka_2 [/itex] and the same goes for Ka1 - these values seem to correspond with the pK_a values for oxalic acid, so I figured it had something to do with it. Then I improvised the equations to get what I wanted (doubtedly correct):

a) C₂O₄^2- + H₃O⁺   HC₂O₄^- + H₂O
b) HC₂O₄^- + H₃O⁺ H₂C₂O₄ + H₂O

Where for a), K corresponds with K_a1^-1, same goes for b) and K_a2

I then arrived at the expression:

K_a1*K_a2*[H₂C₂O₄] = [H₃O⁺]²[C₂O₄^2-]

But I'm confused what [H₂C₂O₄] equals? Since now I got 2 unknowns, including the [C₂O₄^2-], which I guess I'm trying to calculate.

[AWK]

If I'm correct, s = [Zn2+] + [C2O42-]

s = [Zn²⁺] = [C₂O₄^2-] + [HC₂O₄^-] + [H₂C₂O₄]

[Borek]

http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-acid-base

See equation 9.13 and think how it can be applied to the system.

And in future please don't hijack threads, start a new one.

[Anthasci]

C_a will equal [itex] \frac {K_(sp)}{[C_2O_4^{2-}]} [/itex] and [A^2-] = [C₂O₄^2-], right?

I'll keep that in mind in the future, sorry

[Borek]

C_a will equal [itex] \frac {K_(sp)}{[C_2O_4^{2-}]} [/itex]

No. C_a is the analytical concentration of the acid, so it is a sum of concentrations of all forms. See AWK post for a hint.

and [A^2-] = [C₂O₄^2-], right?

Yes.

Beware: page I linked to uses overall dissociation constants, not stepwise ones.