Question

What volume (in mL) of 1.0 M HCl is required to prepare 500 mL of pH 7.76 buffer using triethanolamine that is 0.020 M in both triethanolamine and triethanolammonium ion?

Answer 1

Kb = 5.8*10^-7

pKb = -log(Kb) = -log(5.8*10^-7) = 6.2365

so:

initial buffer is 0.02 M of both, base + conjugate

for a basic buffer change:

pOH = pKb + log(conjugate/base)

pH = 7.76 so

pOH = 14-pH = 14-7.76 = 6.24

6.24 = 6.236+ log(conjugate/base)

conjugate/base = 10^(6.24 -6.236) = 1.0092

note that the ratio is kpt

0.02 mol of triethanolamine

requires 0.01 mol of HCl

so

0.01/1 * 1000 = 1 mL of acid is required to turn 0.02 M of base to 0.01 M of base + 0.01 M of conjguate