Question

In a buffer solution composed of .5 M CH3COOH, 4.5*10^-5 M of H+, and .2M CH3COO- with a pH of 4.35, how many moles of NaOH must be added to raise the pH to 4.5? The Ka for CH3COOH is 1.8*10^-5. (If possible, please do not use the Henderson-Hasselbach equation [pH=pKa+log(moles base/moles acid)] because my professor doesn't want us to use it)

Answer 1

The reaction is

CH3COOH (aq) <---> CH3COO-(aq) + H+ (aq)

initially CH3COOH = 0.5M , [H+] = 4.5x10^-5 M , [CH3COO-] = 0.2 M

Volume = 1L ( since not mentioned we will asusme 1L)

Moles = Mx V , hence CH3COOH moles= 0.5 , H+ moles= 4.5x10^-5 , CH3COO- moles = 0.2

Let NaOH moles be m, NaOH gives OH- which reacts with CH3COOH to give CH3COO-

hence at equilibrium CH3COOH moles = 0.5-X , CH3COO- moles = 0.2+X ,

given pH = 4.5 which means [H+] = 10^-4.5 = 3.162 x 10^-5

Ka of CH3COOH [CH3COO-][H+]/[CH3COOh]

1.8 x 10^-5 = (0.2+X) (3.162x10^-5) / ( 0.5-X)

0.9 x 10^-5 -1.8x10^-5 X = 6.32455 x 10^-6 + 3.162x10^-5 X

X = 0.0539

we can recheck this with Henderson eq

pH = pka + log [CH3COO-]/[CH3COOH]       where pka = -log Ka = -log ( 1.8 x 10^-5 = 4.745)

pH = 4.745 + log ( 0.2+0.0539) / ( 0.5-0.0539)

= 4.5