[Yurij]

Hello!

I apologise that my first post here is a question and not an answer but I really need help solving this:

Prepare 1L of buffer with a pH value of 6 using 0,5M Na₃AsO₄ and 0,4M HCl.
pK_a1=2,2  pK_a2=6,9 and pK_a3= 11,4

I know (or think I do) I need to use pKa2 since it is closest to the desired pH. My idea was that volume of HCl is HCl needed to transform A3- to HA2-  plus volume of HCl which transforms some of the latter to HA- The difference beween 1 and what I'd get is volume of Na₃AsO₄.

Can someone please tell me if I'm thinking right and please please provide a hint or two about what to do next.

responses much appreciated.

[AWK]

From pH you can calculate a ratio of [H₂ASO₄^-]/[HAsO₄^2-]. To solve the problem you need also one of these four three: final concentration of buffer, volume of salt, volume of acid used or the information that sum of volumes (salt and acid) is equal 1 L. In every case you should set system of two equation with two variables.

[Yurij]

Ok so you get

K_a2/[H⁺] =[HA^2-]/[H2A^-] = 0,11

but I still do not see what the second system should be. I know V(HCl) + V(A^3-) (so: V(A^3-) =1L - V(HCl) but I cannot come up with a suitable equation. Please help some more.

Thanks in advance

[AWK]

Now you have a first from three equations:
[H₂ASO₄^-]/[HAsO₄^2-]=x/y=0.11
Now express moles of AsO₄^3- through x, y and unknown volume (V₁) of this salt (c=0.5) used for buffer preparation.
In the same manner express moles of HCl through x, y and unknown volume (1-V₁) of HCl (c=0.4)
You have 3 equation with 3 variables.

Remember about stoichiometry of formation HAsO₄^2-!
Show you calculations on the forum, it is a nice problem.

[Yurij]

Darn. I'm still so confused. The way I see this is that this is two step system, with the first step being transforming all of initial salt into HAsO42-

AsO₄^3-         +        HCl -->          HAsO4^2- +       NaCl

and then transforming some of the latter the latter into H₂AsO₄^-.

HAsO₄^2-   +        HCl -->  H₂AsO₄^-

And now I Always think I'm adding two separate volumes of HCl and this is making it impossible to calculate anything since it is giving me another variable.

Number of moles of AsO₄^3- is equal to the number of moles you'd get if you multiplied equilibrium concentrations from K_a2 with 1L and then added them up or am I wrong? Where do I use volume in the equation?

I'm really sorry to be a bother but this problem is making my brains boil


EDIT: Never mind, I think got it. I will write the calculations sometime today

[Yurij]

I apologise for double post but I couldn't edit in time. I would also like to apologise because I will use trianguar brackets for equilibrium concentration because of convenience on my part (keyboard).

Here's how I did it:

#0.5:  V_HCl + V_A^3-  =  1L

#1: K_a2/<H⁺> = <HA^2->/<H₂A^-> = 0.11

#2: Same goes for number of moles of each of above components in final solution.

#3: 0.5M x V_A^3-  =  n(H₂A^-) + n_e(HA^2-)

#4  0.4M x V_HCl = 2n(H₂A^-) + n_e (HA^2-)

Subtracting 3 from 4 gets you:

#5   n(H₂A^-)  =  0.4M x V_HCl - 0.5M x V_A^3-

Inserting 5 into 3 gets you

#6 n_e(HA^2-) = 1M x V_A^3-  -  0.4M x V_HCl


Now insert both 5 and 6 into first equation, the ratio of moles is the same as the equilibrium concentration ratio since you divide both of the latter with 1 to get number of moles.

After rearranging you get

#7   1.055M x V_A^3- = 0,444M x V_HCl

From here on I believe it's pretty obvious.

Thanks again to AWK, your help was invaluable. I'm really grateful.