I have 2 questions which I don't comprehend at all. I did attempt the questions but obviously I do not know what I am doing. Will help if someone could guide me step-by-step.
1. 20.0cm3 of 0.10 mol dm-3 aq Na2CO3 was titrated witih 0.20 mol dm-3 aqueous HCl.
The 2 base dissociatin constants of the carbonate ion are Kb1 = 2.08 X 10^-3 mol dm-3
Kb2 = 2.32 X 10^-8 mol dm-3
Calculate pH of solution when 0cm3, 5cm3, 15cm3,20cm3 and 25 cm3 of HCl was added.
Sketch the titration curve labeling the points and indicating buffer regions (i should be able to do this part PROVIDED i can understand the previous parts above, lol) so yeah, if equations can be written to explain to me what's happening, that will help. I also do not know why we are given 2 Kb values.
2. A biochemist needs to prepare a CH3COOH/CH3COO-Na+ buffer sollution to stimulate growth of a particular fungus.
In order to investigate the composition of this buffer of pH 4.16, he titrated 25.0cm3 of this buffer solution with 1.0mol dm-3 aq NaOH.
Upon addition of 3x cm3 of aq NaOH, buffer solutionn becomes most effective. Equivalence point attained when 8x cm3 of aq NaOH added.
(a) Show that ratio of [CH3CooH]: [CH3COO-Na+] is 4:1 in orginal buffer solution
(b) Determine Ka of CH3COOH.
Do guide me if anyone could... I want to solve these questions asap :)
Thanks all!
kindly clear the big chunk of HTMLs..
Due to a chunk of unknown HTMLs appearing in my thread... I am reposting this question again.
I have 2 questions which I don't comprehend at all. I did attempt the questions but obviously I do not know what I am doing. Will help if someone could guide me step-by-step.
1. 20.0cm3 of 0.10 mol dm-3 aq Na2CO3 was titrated witih 0.20 mol dm-3 aqueous HCl.
The 2 base dissociatin constants of the carbonate ion are Kb1 = 2.08 X 10^-3 mol dm-3
Kb2 = 2.32 X 10^-8 mol dm-3
Calculate pH of solution when 0cm3, 5cm3, 15cm3,20cm3 and 25 cm3 of HCl was added.
Sketch the titration curve labeling the points and indicating buffer regions (i should be able to do this part PROVIDED i can understand the previous parts above, lol) so yeah, if equations can be written to explain to me what's happening, that will help. I also do not know why we are given 2 Kb values.
2. A biochemist needs to prepare a CH3COOH/CH3COO-Na+ buffer sollution to stimulate growth of a particular fungus.
In order to investigate the composition of this buffer of pH 4.16, he titrated 25.0cm3 of this buffer solution with 1.0mol dm-3 aq NaOH.
Upon addition of 3x cm3 of aq NaOH, buffer solutionn becomes most effective. Equivalence point attained when 8x cm3 of aq NaOH added.
(a) Show that ratio of [CH3CooH]: [CH3COO-Na+] is 4:1 in orginal buffer solution
(b) Determine Ka of CH3COOH.
Do guide me if anyone could... I want to solve these questions asap :)
Thanks all!
A buffer solution is titrated with NaOH(aq). Upon addition of 3 volumes of NaOH(aq), the buffer solution becomes most effective. Equivalence point is attained when 8 volumes of NaOH(aq) is added.
a) Calculate the ratio of the molarities of both members of the conjugate acid-base pair of the buffer solution, before the NaOH(aq) is added.
b) Given that the original pH of the buffer solution (ie. before NaOH(aq) is added) is 4.16, calculate the proton dissociation constant of the acidic component of the buffer.
Solution :
a) At maximum buffer capacity :
HA + OH- ---> A- + H2O
Initial (mol) 8y 3y x-8y n.a.
Change (mol) -3y -3y +3y n.a
Final (mol) 5y 0 x-5y n.a.
At equivalence point :
HA + OH- ---> A- + H2O
Initial (mol) 8y 8y x-8y n.a.
Change (mol) -8y -8y +8y n.a.
Final (mol) 0 0 x n.a.
First, complete the ICF table for equivalence point. Next, complete the ICF table for maximum buffer capacity, bearing in mind that the initial moles of HA and A- are the same (for both tables).
Since at maximum buffer capacity, [HA] = [A-], this implies 5y = x-5y and hence x = 10y.
Substituting x = 10y into the initial moles of A-, we have moles of A- = 2y.
Therefore, the ratio of the amounts of both members of the conjugate acid-base pair present in the buffer (before NaOH(aq) is added), is 8y HA : 2y A-, ie. 4 HA : 1 A-
b) Using the Henderson-Hasselbalch equation, we have
pH = pKa + log ( [base] / [acid] )
4.16 = pKa + log (1/4)
pKa = 4.762
Ka = 1.73 x 10^-5
Originally posted by Audi:1. 20.0cm3 of 0.10 mol dm-3 aq Na2CO3 was titrated witih 0.20 mol dm-3 aqueous HCl.
The 2 base dissociatin constants of the carbonate ion are Kb1 = 2.08 X 10^-3 mol dm-3
Kb2 = 2.32 X 10^-8 mol dm-3
Calculate pH of solution when 0cm3, 5cm3, 15cm3,20cm3 and 25 cm3 of HCl was added.
Sketch the titration curve labeling the points and indicating buffer regions.
Do ICF (moles) tables for the different volumes of NaOH(aq) added. You'll find that 0cm3 and 5cm3 are before the 1st equivalence point, while 15cm3 is between 1st and 2nd equivalence points, and 20cm3 is for 2nd equivalence point, and 25cm3 is past the 2nd equivalence point.
To find pH at these various points :
0cm3 : Use Kb of CO3 2-
5cm3 : Use Henderson-Hasselbalch with [CO3 2-] / [HCO3 -]
15cm3 : Use Henderson-Hasselbalch with [HCO3 -] / [H2CO3]
20cm3 : Use Ka of H2CO3. Be mindful of the volume change.
25cm3 : Calculate [H+] contributed from HCl only, and thus calculate pH. Write a statement to explain that because HCl is a strong acid and H2CO3 is a weak acid, the protons dissociated from HCl suppresses the proton dissociation from the weak acid H2CO3 (as predicted by Le Chatelier's principle), and thus we only need to consider [H+] from HCl. Be mindful of the volume change.