Question:

Question

50cm3 of 0.4 M Ba(OH)2 and 50cm3 of HA of unknown concentration is added together.At 25°C,pH of the solution is 9.With appropriate assumptions calculate the [HA] and ka of HA.

My thoughts on the question:

There are more than one approach which came to mind...I'll try to type them as briefly as I can and understandable by the reader.

Reactions:

1)Ba(OH)2+ 2HA--->BaA2+ 2H2O
2)BaA2-->Ba2+ + 2 A-
3)A^- <===> HA + OH-
4)HA<==> H+ + A-

Let the [HA]=c moldm-3

When they are mixed,volumes become double value and concentrations of them become half of the initial values.

If we neglect the ion concentrations we get from the hydrolysis, from (4) [H+]= (10)^-9 M.But we can't simply say [A-] = (10)^-9 M if BaA2 dissociate completely. (Does it completely dissociate????)

If not from (4) , [H+]=[A-]= (10)^-9 M and the [ HA]=(c/2 - (10)^-9 ) M
So ka=( (10)^-9 )^2/[(c/2) - (10)^-9 ]
But we have two unknowns ka and c.So what should be the other equation/relationship ????

-another approach that came to my mind-

Considering the stoichiometry of (1),if we take HA as the limited reactant and it reacts completely and then finding BaA2 moles/concentration formed etc... and finding ka...
(Can we actually take HA reacts completely. It really doesn't sound sensible to me)

And if we consider the hydrolysis,
Hydrolysis constant =(kW)/(ka)
So ka= kW/kh
KW=(10)^-14 ,but kh is not given...

Or is there any other approach???

Answer 1

You don't post easy questions. Thanks for sharing your thoughts. That helps. Here are my thoughts/questions.
1. The problem says make appropriate assumptions and calculate (HA) and Ka. I'm not sure what "appropriate" means so I went through the possibilities. There MAY be others; if so please point them out to me. Here are mine:

a. IF HA is the limiting reagent (LR)(we can assume that I guess), then the pH of the final solution is determined by the excess (xs) Ba(OH)2.

b. IF Ba(OH)2 is the LR instead of HA, then HA is in xs and the solution will be a buffered solution with the salt and xs acid. The pH will be determined by the Henderson-Hasselbalch equation.

c. IF neither Ba(OH)2 nor HA are in xs (they exactly neutralize each other), that means there is only the salt at the end and the pH will be determined by the hydrolysis of the salt.

I started at the bottom and worked my way to the top. I don't think it can be c. If hydrolysis, and we let c stand for the equilibrium concentrations, then
.....A^- + HOH ==> HA + OH^-
I....c..............0....0
C....-x.............x....x
E....c-x............x....x

Kb for A^- = (Kw/Ka for HA) = (x)(x)/(c-x). You know Kw and x but not Ka nor Ka so I ruled this one out because of two unknowns.

b. I ruled out b for the same reason.
pH = pKa + log (base)/(acid)
You know pH and (base) but not pKa nor (acid). Two unknowns again. That left me with just a as a possibility. So I went with that. Here are my thoughts.

...Ba(OH)2 + 2HA ==> BaA2 + 2H2O
pH after mixing is 9 so pOH must be 5 and (OH^-) is 1E-5. In 100 mL that will be 1E-5 * (100/1000) = ? mols OH. Convert that to mmols, then convert mmols OH^- to mmols Ba(OH)2.
You had 100*0.2 = 20 mmols Ba(OH)2 to start. You used initial Ba(OH)2 - amount left and twice that will be amount HA used. Since HA is the LR that makes that the (HA). Then set up
Ka = (H^+)(A^-)/(HA). You know H^+ and A^- and HA so you can calculate Ka. I don't think I've missed anything. I've not put in all the numbers but you can do that. If you have questions, please show your work.

Answer 2

Your thoughts seems to be fine and logical!
I have a few questions

1)At start shouldn't we take the moles of Ba(OH)2=(0.4)*50*(10)^-3 moles ,considering the concentrations becoming half of the initial values as the volume becomes double value.(Or at start point we should just use the initial concentrations to find the moles???)

2)Ba(OH)2 moles in 100cm3= half of the OH- moles in 100cm3,which is equal to 5*1E-7 moles

So the HA moles used= 2*{[0.2*50*(10)^2*(10)^-3] - [5*(10)^-7]} ,which is a very close value to 2*(0.2)*50*(10)^2*(10)^-3.
In that case,if we take the moles of HA used approximately as = 2*(0.2)*50*(10)^2*(10)^-3=2 mol(????)
(Did I made a mistake??)

So as I considered the initial concentration of HA=c moldm-3,
moles of HA used(as HA being the LR)= [(c/2)*50*(10)^-3]=2.
Then I got a (ridiculous/unreal) value for c.
(Where did I made a mistake???)

Answer 3

For the second question I found my mistake.
I've calculated initial Ba(OH)2 moles as (0.2)*50*(10)^2*(10)^-3.It should be (0.2)*50*(10)^-3

Then the HA moles used approximately equal to=2*(0.2)*50*(10)^-3= (c/2)*50*(10)^-3
So c=(2/25)*(10)^2*(10)^(-3)=(2/25)*(10)=0.8,twice as the concentration of Ba(OH)2,which sounds sensible to me

Answer 4

From your last post, initial mols Ba(OH)2 = 0.4 x 50E-3 or if you want to use the 0.2, because the 50 cc was diluted to 100 cc, then mols Ba(OH)2 = 0.2 x 100E-3. Either way you have same mols which is why I like to use the initial mols Ba(OH)2 as 0.4*50E-3, work through the problem with mols that are used and convert to the diluted values at the end when the question is for concns Note that diluting changes the concentrations but the number of moles is not changed by dilution. Just to throw out some numbers if we start with 20 millimols Ba(OH)2 (0.02 mols--I like mmols because I get lost with all those zeros), the amount HA used is about 38 mmols (0.038 mols) but I didn't check my math on that.

Answer 5

If you follow up I suggest you post new at the top; otherwise, this will be so far down the list that it will be difficult to find.

Answer 6

From next time I'll do so.Sorry for the inconvenience caused and thank you for your help.