Question

(a) Find the concentration of H+, HCO3
- and CO3
2-, in a 0.01M solution of carbonic acid
if the pH of this is 4.18.
Ka1(H2CO3)=4.45 10–7 andKa2 =4.69 10–11
(b).Calculate the pH at the equivalence point of the titration between 0.1M CH3-
COOH ( 25 ml) with 0.05 M NaOH. Ka (CH3COOH) = 1.8 10–5.

Answer 1

H2CO3 ==> H^+ + HCO3^-
HCO3^- ==> H^+ + CO3^2-

k1 = (H^+)(HCO3^-)/(H2CO3)
k2 = (H^+)(CO3^2-)/(HCO3^-)

For a. You know pH. pH = -log(H^+). Solve for (H^+)
Use k1, You know H^+ and H2CO3, substitute and solve for HCO3^-.
Use k2. You know H^+ and HCO3^2-. Substitute
and solve for CO3^2-

For b. Let's call CH3COOH a simpler HAc.
..............HAc + NaOH ===> NaAc + H2O
millimols HAc = mL x M = 25 x 0.1 = 2.5
So you will need 2.5 millimols NaOH. M = mmols/mL and mL NaOH needed is mL = mmols/M = 2.5/0.05 = 50 mL of the NaOH. What is the concentration of NaAc at the equivalence point? It is (NaAc) = mmols/mL = 2.5/75 mL = approx 0.033. The pH at the equiv pt is determined by the hydrolysis of the Ac^-.
.............Ac^- + HOH ==> HAc + OH^-
I............0.033......................0.........0
C..........-x.............................x..........x
E...........0.033-x...................x...........x

Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/*0.033-x)
Plug in Kw, Ka for HAc (given in the problem), solve for x = (OH^-) and convert to pH.
Check those numbers and not tht the 0.033 is a estimate.

Post your work if you get stuck.

Answer 2

Ye kya chutiya panti mari hai

Answer 3

Banchod kya kr diya tune laude... Randi