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An acetic acid/ sodium acetate buffer solution similar to the one you made in the lab...

An acetic acid/ sodium acetate buffer solution similar to the one you made in the lab was prepared using the following components: 3.46 g of NaC2H3O2∙3H2O (FW. 136 g/mol) 9.0 mL of 3.0 M HC2H3O2 55.0 mL of water If you take half of this solution and add 2 mL of 1.00 M NaOH to it, then what is the pH of this new solution?

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Answer #1

this is a buffer since

HC2H3O2 = weak acid

NaC2H3O2 = conjugate base

are present

so we must model this as a buffer, use henderson hasselbach equations

pH = pKa + log(base/acid)

pH = pKa + log(NaC2H3O2 /HC2H3O2 )

pKa = 4.75 for HC2H3O2

initially

mol of NaC2H3O2 = mass/MW = 3.46/136 = 0.0254

mol of HC2H3O2 = MV = 9*3*10^-3 = 0.027

after adding

mol ofNaOH = MV = 2*1 = 2*10^-3 mol

mol of NaC2H3O2 = 0.0254 -  2*10^-3 = 0.0234

mol of HC2H3O2 = 0.027+ 2*10^-3 = 0.029

now..

pH = pKa + log(NaC2H3O2 /HC2H3O2 )

pH = 4.75+ log(0.0234/0.029)

pH = 4.6568

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