Finding pH of an Acid with Added OH-: Ka and the Quadratic Equation

[fish]

__________HA <------> H+ + A-
initial (mol) .1 ________ 0 ___ 0
change __ -x _______ +x ___ +x
final _____.1-x _______ x ___ x

Ka = 10^-8
find pH of .1 mol acid
since ka X 100 = 10^-6 < .1 can use initial concentration: .1-x = .1

Ka = [H+][A-]/[HA]
10^-8=x^2/.1
x=[H+]= 3.16x10^-5
pH = 4.5

now add .01 mols of OH- and see how it changes the pH

Ka=x[3.16x10^-5 + .01]/(.1 - .01)
10^-8 = .01x/.09
x=9x10^-8

-log(9x10^-8) = pH of 7 (also a pOH of 7)

what is the 9x10^-8 ion concentration representing, the [OH-] or [H+] ions?

if it's the [OH-] then shouldn't it be 14-pOH = pH
14-7=7

is this formula the best way find the pH in this case?
Ka=x([H+] + [OH-])/([HA] - [OH-])

 

[switch]

A little confused- can you give some indication as to volume/amount. I don't know if I am reading that right but you seem to have derrived pH from moles rather than mol dm^-3- I'm a little tired at the moment so I don't know though.

 

[iluvsr20s]

now add .01 mols of OH- and see how it changes the pH

Ka=x[3.16x10^-5 + .01]/(.1 - .01)
10^-8 = .01x/.09
x=9x10^-8

I'm not quite sure what you did here could you post the actual problem and then explain what you did above

 

[iluvsr20s]

what is the 9x10^-8 ion concentration representing, the [OH-] or [H+] ions?

to answer this question the 9*10^-8 represents both the OH- and the H+ since your pH is 7 those two concentrations have to be equal

 

[iluvsr20s]

Okay i got it see if you follow me

your going to take 14-4.5=pOH then take 10^-pOH=[OH-] + the [.01] your adding, that equals .01 then take the -log.01=pOH 14-pOH=pH