Question : ka1 and ka2 values of H2X is given 1*(10)^-5 M and 1*(10)^-9 M.50cm^3 of 0.2 M KOH and 50 cm^3 of H2X of unknown concentration is mixed together and the pH of the final solution is given as 7.
1)Find the concentration of H2X
2)Find the pH of the initial H2X solution(before being mixed with KOH)?
Work:
I took the concentration of H2X as c(moldm-3)
As KOH is a strong base I considered that it reacts completely with H2X.
When they are mixed together their volumes get double so their concentrations become half of the initial values.
H2X(aq) + KOH(aq) -----> KHX(aq) + H2O(l)
After the reaction the concentrations of H2X and KHX are respectively [ (C/2) - 0.1)] and (0.1) M
When applied to ka1 equation we get the C= 1*(10)^-3 M
2) as ka2 is much smaller than ka1 we can neglect ka2.
H2X <=====> H+ + HX- (all are in aqueous form)
After ionization , by applying values in ka1 equation again,we get pH of the initial solution as 4.
Am I correct in the process?
Yes
I Believe so
Thank you very much Emily! And if we titrate 25cm^3 of the final solution with 0.1 M HCl,using Me.O as the indicator,is there any reaction taking place?
I think no as they are both acids(KOH is reacted completely) and the concentrations of OH- we get from hydrolysis of water is a very low value (1*10^-7 M) and also the Concentrations of other ions such as KHS X2- are too small to react with HCl of 0.1 M
What do you think?
I think you are VERY close but not quite there.
For 1,I think you substituted into Ka1 to solve for C but in reality I think you should have used the Henderson-Hasselbalch equation because you have formed a buffer with KHX and H2X.
pH = pKa1 + log (base)/(acid)
7 = 5 + log(0.1/c/2)
7 = 5 + log(0.2/c) and go from there. I get c = 0.002 but you should confirm that.
For 2, then, since C is not 0.001 M you need to recalculate pH. I didn't calculate it but you MAY need to use a quadratic since 0.002 and ka1 are close. Now for the follow up you asked Emily. YES, if you add HCl to KHX it will titrate it to the MO end point.
KHX + HCl ==> HCl + H2X
In fact, that's how you titrate Na2CO3. You add phenolphthalein and titrate to the first end point(you have NaHCO3 there), then you add methyl orange indicator and titrate to the second end point where you form H2O and CO2 (from H2CO3). You can determine mixtures of NaHCO3 and Na2CO3 this way. If you have follow up questions I suggest you post a NEW question to get it closer to the top.
Thank you very much DrBob222.I totally forgot about the Henderson equation!
I think you cleared out all my doubts.Thank you very much for the help..
Now I'm having doubts. As I think about it I'm suspicious that 0.002 just doesn't sound like a number that would be right. And if you look at 50 cc of 0.002 M H2X reacting with 50 cc of 0.2M KOH,it doesn't make sense that there is KOH left over sine an excess of KOH certainly wouldn't leave a pH of 7. then I think this requires more thought. I believe Im overlooking something here so don't take this answer to the bank. I went to bed and got back up to write this. I'll rethink this tomorrow. It's almost 3 A.M here and I need the sleep. _=)
Thank you very much for your concern on this question in the first place.
And I forgot to include that It is given that the temperature is 25°C.
Again I'm also having doubts.I will wait for your answer and in the mean time I'll try this again.