[PopcornMuffin]

How many moles of NaOH must be added to 0.550 L of 0.185 mol/L HC7H5O2(aq) to obtain a solution with pH=3.9? Assume the temperature is 25oC. (For HC7H5O2, pKa=4.21 at 25oC).

Attempt:
3.9 = -log[Needed H+]
Needed [H+] = 1.25893E-4 = y

HC7H5O2 + H2O --> C7H5O2- + H3O+
pKa = -logKa
Ka = 6.166E-5

Ka = [C7H5O2-][H+] / [HC7H5O2]
6.166E-5 = x^2 / (0.185 - x)
x = 3.3346737E-3 M

(0.55L)(x) - (0.55L)(y) = 0.001771463 mol
Therefore 1.77E-3 mol of NaOH is needed...



  Where did I go wrong?

[Rabn]

Where you went wrong is in what you call needed [H+].  Find out how many moles of H+ you need to remove from the solution in order to attain the desired pH and work from there.

[PopcornMuffin]

Isn't 0.55(x-y) in the above attempt the amount of moles H+ I need to remove and therefore also the answer since thats the amount of mol NaOH I need to neutralize the H+?

[AWK]

0.185 is a concentration, not moles. Hence x is the concentration. Calculate moles assuming unchanged volume.

[Rabn]

Maybe you should try using henderson-hasselbalch...

[AWK]

Ka = [C7H5O2-][H+] / [HC7H5O2]
66.166E-5 = x^2 / (0.185 - x)

You know [H+], then use equation

6.166E-5 = [H+]x / (0.185 - x)

[PopcornMuffin]

Ka = [C7H5O2-][H+] / [HC7H5O2]
66.166E-5 = x^2 / (0.185 - x)

You know [H+], then use equation

6.166E-5 = [H+]x / (0.185 - x)

Ah I see, makes sense now. Thanks for the help.

6.166E-5 = (10^-3.9)(x) / (0.185 -x)
x = 0.060820823
(.55L)(x) = 0.033451453 mol
Therefore 3.35E-2 mol of NaOH needed?