[Matt17]

Yes I've made a mistake in calculations, K_b was wrong.
K_b = 6,25 * 10^-10
So I'm counting x again.
[tex]6,25*10^{-10} = \frac{x^2}{0,05-x}[/tex]
[tex]x^2 + 6,25*10^{-10}x - 3,125*10^{-11} = 0[/tex]
[tex]x = [OH^{-}] = 5,59*10^{-6} \frac{mol}{dm^3} [/tex]
And of course I know that it's not the end:
[tex]pOH = - log 5,59*10^{-6}[/tex]
[tex]pOH = 5,25 [/tex]
[tex]pH = 14 - 5,25 = 8,75 [/tex]
Which means that the solution will be basic.
Is that correct?

I also have 2 more questions:
1. Hypothtically if I included ioinic strength in calculations, would the error be small enough not to affect the final answer - that the solution is basic? And if not, than in my answer can I simply write that the influence of NaCl is so small that it can be omitted in calculations?
2. I changed the original question a little bit because I didn't know how to translate one phrase to English but I'll try to do it so that I can be sure that everything is correct: The original qustion didn't say that there were 100 cm³ of both solutions but said that: 'Concentrations of all electrolyte solutions are 0,1 mol/dm³. Solutions of CH₃COONa and NaCl are mixed in the volume ratio of 1:1 (by volume ratio I mean that both solutions must have the same size eg. 100 cm³, 1 dm³ or whatever). So did I use the concentration of 0,05 in calculations correctly?

Sorry for such a long post.

[Borek]

[tex]pH = 14 - 5,25 = 8,75 [/tex]
Which means that the solution will be basic.
Is that correct?

Without checking numbers - yes, it is definitely in the right ballpark now.

1. Hypothtically if I included ioinic strength in calculations, would the error be small enough not to affect the final answer - that the solution is basic? And if not, than in my answer can I simply write that the influence of NaCl is so small that it can be omitted in calculations?

In this particular case ionic strength effect would be that the pH is about 0.1 pH unit lower.

2. I changed the original question a little bit because I didn't know how to translate one phrase to English but I'll try to do it so that I can be sure that everything is correct: The original qustion didn't say that there were 100 cm³ of both solutions but said that: 'Concentrations of all electrolyte solutions are 0,1 mol/dm³. Solutions of CH₃COONa and NaCl are mixed in the volume ratio of 1:1 (by volume ratio I mean that both solutions must have the same size eg. 100 cm³, 1 dm³ or whatever). So did I use the concentration of 0,05 in calculations correctly?

Why don't you just quote the original wording? 

[Matt17]

Why don't you just quote the original wording? 

Because it's not in English. But as far as I can see you are Polish too so I'll write it in Polish (I hope I won't be punished for this). 'Stężenia roztworów elektrolitów wynoszą 0,1 mol/dm³. Roztwory CH₃COONa i NaCl są zmieszane w stosunku objętościowym 1:1.
That's it.

[Borek]

Whoever worded the problem, wanted you to be confused.

But yes, that would mean you start with 0.1 M solutions and you mix them 1:1, ending with 0.05 M.

[Matt17]

You are very helpful! Thank you very much! I've learnt a lot!

But I have a very similar question though it looks easier:

Solutions of CuSO₄ and Na₂SO₄ are mixed together in 1:1 proportions (and concentration of both of them is again 0,1 mol/dm³). The question is the same: is this mixture acidic, basic, neutral or close to nuetral? There aren't any constants given, so I assume that I don't have to count anything.

Na₂SO₄ only dissociates but doesn't hydrolize. And there aren't any reactions between CuSO₄ and Na₂SO₄.
CuSO₄ hydrolizes:
Cu²⁺ + SO₄^2- + H₂O  (CuOH)⁺ + HSO₄^-
(CuOH)⁺ + H₂O  Cu(OH)₂ + H⁺
Which means that the mixture is acidic.

Am I correct?

[Borek]

CuSO₄ solution should be definitely slightly acidic. No idea how much though.

[Matt17]

Thank you! I'd be grateful if you could help me with one more problem.
There are solutions of KF and HCl mixed in proportions of 2:1 (so we can assume that the proportions are the same but concentration of KF is 0,2 mol/dm³ and concentration of HCl is 0,1 mol/dm³). The same question. Constants: K_a(HF) = 6*10^-4 .

What I am sure of is that:
KF hydrolizes:
F^- + H₂0  HF + OH^-

and
The dissociation of HF is:
HF + H₂O  H₃O⁺ + F^-

And now I don't know if HCl will react with F^- or maybe OH^-. And then I don't what would be the next step, which ions and reactions to consider in calculations.

I'd be very glad if you could help me with this one, too.

[Borek]

http://www.chembuddy.com/?left=buffers&right=composition-calculation

See 5^th problem (the one about HCl and sodium acetate).

[AWK]

pK_a of hydrated copper(II) is close to 10^-8.
Note - if you mix your compound with salt that increase ionic strength then you should correct your pH result for ionic strength, though for 0.05 M CuSO₄ + 0.05 M Na₂SO₄ (I~ 0.3) this is quite difficult.
Try for example aq_solution free program
http://www.acadsoft.co.uk/download/Aq_solutions.zip

[Matt17]

I think I got it:

The reaction:
KF + HCl  HF + KCl

And assuming that we have 1 dm³ of 0,2 mol/dm³ KF and 1 dm³ of 0,1 mol/dm³ HCl
[tex][HF] = \frac{0,1}{2} = 0,05[/tex]
[tex][F^-] = \frac{0,2 - 0,1}{2} = 0,05[/tex]
[tex]\frac{[HF]}{[F^-]} = 1[/tex]
Which I think means that
[tex][H^+] = K_a * 1[/tex]

I hope that's correct?

[Borek]

KF + HCl  HF + KCl

Go net ionic, and it is not  but 

And assuming that we have 1 dm³ of 0,2 mol/dm³ KF and 1 dm³ of 0,1 mol/dm³ HCl

Not necessarily follows from the wording of the problem that you posted earlier. Doesn't mean it is wrong.

Which I think means that
[tex][H^+] = K_a * 1[/tex]

Looks OK. For a solution where [A^-]= [HA] pH=pKa.

For next problems please start new threads. And an English lesson: you are not counting, you are calculating. In Polish there is no difference, but counting refers to finding number of objects (like you have some eggs and to find out how many are there you count them), while calculating refers to finding a number (to find out how much is 3*5 you calculate). Jest pewna analogia do różnicy między liczbą i ilością - liczba dotyczy obiektów przeliczalnych, jak siedem długopisów (counted), ilość dotyczy obiektów nieprzeliczalnych, na przykład ilość soli w roztworze (którą możesz policzyć - n=C×V - calculate).

[Matt17]

Thank you very much. And thank you for an English lesson - very useful one.
And of course, if I have any more problems with pH, I'll start a new thread.

[Matt17]

I have one more question. Why don't we have to consider hydrolysis of KF in this example? Is that because it is a buffer?

[Borek]

I have one more question. Why don't we have to consider hydrolysis of KF in this example? Is that because it is a buffer?

In a way - yes. It can be easily shown that the error we make is negligible, compare:

http://www.chembuddy.com/?left=buffers&right=with-ICE-table