Experimentally, acetate buffer HC2H3O2-C2H3O2^- can be prepared in different way

Question

Experimentally, acetate buffer HC2H3O2-C2H3O2^- can be prepared in different ways. a. Calculate the approximate pH of an acetate buffer solution prepared by mixing equal volumes of 0.10 M HC2H3O2 and 0.10 M NaC2H3O2 (Ka, HC2H3O2 = 1.8 x 10^-5). Show your work. b. If an acetate buffer solution was going to be prepared by neutralizing HC2H3O2 with 0.10 M NaOH, what volume (in mL) of 0.10 M NaOH would need to be added to 10.0 mL of 0.10 M HC2H3O2 to prepare a solution with pH = 5.5 ? Show your work.

Explanation / Answer

(a) [NaC2H3O2] = [HC2H3O2] = 0.10/2 = 0.05 M

pH = pKa + log([NaC2H3O2]/[HC2H3O2])

= -log Ka + log([NaC2H3O2]/[HC2H3O2])

= -log(1.8 x 10^(-5)) + log(0.05/0.05)

= 4.74


(b) Initial moles of HC2H3O2 = 10.0/1000 x 0.10 = 0.001 mol

Let a be the volume of NaOH added

Moles of NaOH added = a/1000 x 0.10 = 0.0001 mol


HC2H3O2 + NaOH => KC2H3O2 + H2O

Moles of HC2H3O2 left = 0.001 - 0.0001a mol

Moles of NaC2H3O2 formed = moles of NaOH added = 0.0001a mol


pH = pKa + log([NaC2H3O2]/[HC2H3O2])

= -log Ka + log(moles of NaC2H3O2/moles of HC2H3O2)


5.5 = -log(1.8 x 10^(-5)) + log(0.0001a/(0.001 - 0.0001a))

log(0.0001a/(0.001 - 0.0001a)) = 0.75527

0.0001a/(0.001 - 0.0001a) = 10^0.75527 = 5.692

0.0006692a = 0.005692

a = 8.51


Volume of NaOH = a = 8.51 mL = 8.5 mL

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