Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: winged17 on April 04, 2015, 01:02:35 PM
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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.
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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)?
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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?
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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
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It was from the NaOH which is the limiting reactant ....or not
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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
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Oh okay, I get it now, thanks a lot! God bless :)