You are asked to prepare 500. mL of a 0.250 M acetate buffer at pH 5.00 using on

Question

You are asked to prepare 500. mL of a 0.250 M acetate buffer at pH 5.00 using only pure acetic acid (MW=60.05 g/mol, pKa=4.76), 3.00 M NaOH, and water. Answer the following questions regarding the preparation of the buffer.

1. How many grams of acetic acid will you need to prepare the 500 mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution.

2. What volume of 3.00 M NaOH must you add to the acetic acid to achieve a buffer with a pH of 5.00 at a final volume of 500 mL? (Ignore activity coefficients.)
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I solved for the grams  of acetic acid needed - 7.51 grams, but I don't know how to approach the second question.

Explanation / Answer

pH = pka + log [sodium acetate]/[acetic acid]

5 = 4.76+ log [sodium acetate]/[acetic acid]

[sodium acetate] = 1.7378 [acetic acid]

we have [acetic acid]+ [sodium acetate] = 0.25 = buffer conc

by solving above two equations we get

[sodium acetate] = 0.1587

[acetic acid] = 0.0913 ,

2) since we have eq acetic acid + NaOH <-----------> sodium acetate + H2O

hence moles of sodium acetate = ( 0.1587 x500/1000) = 0.07935

we have 3 M soln and we take V amount and dilute to 500 ml

now 3 x V = 0.1587 x 500

V = 26.45 ml hence 26.45 ml NaOH required

1) now acetic acid moles = 0.0913 x0.5 = 0.04565

but acetic acid moles = initial moles of acetic acid - NaOH moles

0.04565 = initial moles - 0.07935

initial acetic acid moles = 0.125

mass of acid/60.05 = 0.125

mass of acetic acid = 7.5 gm

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