- #1
pakmingki
- 93
- 1
im looking at some review problems my teacher gave me today, and i can pretty much do most of them. however, there is this one i really have no idea what to do. Its a 2 part problem.
1a.What is the pH of a 1000 ml of a 0.1 M acetic acid solution, given pKa = 4.8?
1b. what is the pH of 1000ml of 0.1 M acetic acid solution after you have added 5 ml of 10.0 M NaOH, assuming net volume change is negligible
so, for the first part, i think i got it wrong. But here is what i did.
First i found the Ka with
pKa = 4.8 = -log[Ka]
10^-4.8 = Ka
Ka = 1.58 x 10^5
so then i wrote out the equilibrium reaction.
C2H4O2 + H20 > C2H3O2 + H30
where everything is aqueous except H20, and the products on the right side are - and + ions, respectively.
so then
Ka = [x][x]/0.1 - x, where is x is the concentration of H30 and C2H3O2.
solving for x gives 1.25 * 10^-3
so, [Hplus] = 1.25 * 10^-3 M
then, i did this step
(1.25 x 10^-3 Hplus ions/L) * 1000ml * 1L/1000ml gives 1.25 x 10^-3 Hplus ions
pH = -log[Hplus] = 2.903
i really need help, I am pretty sure i made some erroneous steps.
and for part b, i really didnt have any idea how to start. I just started by finding the number of mols of OHminus, but i don't think that does anything.
So, can someone help me do both parts?
thanks
1a.What is the pH of a 1000 ml of a 0.1 M acetic acid solution, given pKa = 4.8?
1b. what is the pH of 1000ml of 0.1 M acetic acid solution after you have added 5 ml of 10.0 M NaOH, assuming net volume change is negligible
so, for the first part, i think i got it wrong. But here is what i did.
First i found the Ka with
pKa = 4.8 = -log[Ka]
10^-4.8 = Ka
Ka = 1.58 x 10^5
so then i wrote out the equilibrium reaction.
C2H4O2 + H20 > C2H3O2 + H30
where everything is aqueous except H20, and the products on the right side are - and + ions, respectively.
so then
Ka = [x][x]/0.1 - x, where is x is the concentration of H30 and C2H3O2.
solving for x gives 1.25 * 10^-3
so, [Hplus] = 1.25 * 10^-3 M
then, i did this step
(1.25 x 10^-3 Hplus ions/L) * 1000ml * 1L/1000ml gives 1.25 x 10^-3 Hplus ions
pH = -log[Hplus] = 2.903
i really need help, I am pretty sure i made some erroneous steps.
and for part b, i really didnt have any idea how to start. I just started by finding the number of mols of OHminus, but i don't think that does anything.
So, can someone help me do both parts?
thanks