I have a chem final tomororrow and have the most difficult time with titration calculations....please help.?

Question

an example problem is Propanoic Acid (CH3CH2COOH) has a Ka of 1.34x10^-5 . A 25.00ml sample of .100 mol L-1 propanocic acid is titrated with .100 mol L-1 NaOH solution, added from a buret.

a) What volume of NaOH is needed to reach the equivalence point?

b) Find the pH when 0.0 mL of NaOH is added.

I really need help with part B i figured out part A for the most part...i need to know where to start and what the heck to do. please help.

Answer 1

part a)
propanoic acid has 1 proton and NaOH has 1 OH so the number of their mols(n) should be equal to reach to eq.point:
concentration=n/volum--->n NaOH=n propanoic acid---->
0.100*25=0.100*volum od NaOH--->V NaOH=25ml

part b)
when 0 ml of Naoh is added the pH is -log[H+] and [H+] is the concentration of H+ in soln:
CH3CH2COOH---->H+ +CH3COO-
Ka=[H+][CH3COO-]/[CH3CH2COOH ]
[H+] and [CH3COO-] have the same concentration for example X,AND concentration of CH3CH2COOH is 0.100 u mentionined in problem substracted X:0.100-X SO we have:
Ka=X^2/0.1-X & PH=-logX :)

Answer 2

You seem to ask what will be the pH before the titration begins, when you still have 0.100M propionic acid. Let propionic acid be called HPr:

Ka = 1.34 x 10^-5 = [H+][Pr-]/[HPr]

[H+] = [Pr-] = x

Then [HPr] = 0.100 - x. But x is so small that we can say that essentially [HPr] = 0.100.

(x)(x)/(0.100) = 1.34 x 10^-5

x^2 = 1.34 x10^-6

x = [H+] = 1.16 x 10^-3

Log[H+] = -2.94

pH = -Log[H+] = 2.94

Answer 3

get off the internet and get back to studying