[winged17]

Calculate the mass of a weak acid, HA,  that must be taken with 50 mL of 1.0 M NaOh to prepare 1 L buffer (pH of 4.5) The HA has a formula mass of 122.12 g/mol and Ka = 6.4 x 10^-5.

[Big-Daddy]

Note first you are safe to work in number of moles, because you have the mass conversion factor for HA and V*c for NaOH.

What are your equations? What approximations can you make (if you need to make any)?

[winged17]

I think I got the answer already. I just used the Henderson-Hasselbalch equation which gave me [ b]/[a]=2.024. I assumed [ b] to be 0.05 mol (from the given NaOH) and then I solved for [a] and got 0.0247 mol. Computing for mass, I arrived at 3.02 grams. Do you think my answer's correct?

[magician4]

Do you think my answer's correct?

no, I don't think so

hint:
though your approach with HH-eqtn. seems sound to me, my impression is, that you forgot to ask one important question: where did those 0.05 moles [A^-] originally come from?



regards

Ingo

[winged17]

It was from the NaOH which is the limiting reactant ....or not

[magician4]

before those 0.05 mol [A^-] did - under the influence of NaOH - become [A^-] , they of course had been HA , too

so, if your equilibrium calls for 0.05 mol A^- and , as counterpart in this buffer, for  0.0247 mol HA, then, of course, you'd have had need for n₀ = 0.05 + 0.0247 mol = 0.0747 mol HA , total ( i.e. before adding NaOH), to begin with.

... meaning m = n * M = 9.12236... g = 9.12 g HA

regards

Ingo

[winged17]

Oh okay, I get it now, thanks a lot! God bless