Question

Explain how you would make 500 mL of a 20 mM citric acid buffer at a pH of 7.0. You must specify the components that you would use and the masses. Explain what will happen if you add .05 mL of 1.0 M NaOH to this solution.

Answer 1

total molarity = mmol*MV = 20*0.500 = 10 mmol of species

if

pH = 7, then

pH = pKa + log(citrate-3/Hcitrate-2)

pKA3 = 6.40

therefore, we need

citric acid = 10 mmol

sodium citrate = 10+10 + x

pH = pKa + log(citrate-3/Hcitrate-2)

7 = 6.40 + log(x / (10-x))

10^(7 - 6.40 ) = x/(10-x)

3.981*10 - 3.981x = x

(3.981+1) x= 3.981*10

x = 3.981*10/(3.981+1) = 7.9923

then

total sodium citrate = 10+10+x = 20+7.9923 = 27.9923 mmol of sodium citrate

mass = mmol*MW = (7.9923*10^-3)(258.06) = 2.062 g of sodium citrate

assume the citric acid is at

0.1 M so..

V of acid = mmol/M = (10)/(0.1) = 100 mL

add 100 mL of a 0.1 M Citric acid solution, then add 2.062 g of Sodium citrate salt, mix well, add water up to V = 500 mL

pH = 7

B)

if we add NaOH --> mmol = MV = 0.05*1 = 0.05 mmol

then,

the pH will increase, since OH- will react with the acid to form more citrat ion