Question

What volumes (in mL) of 1.0 M triethanolamine and 1.0M HNO3 is required to prepare 1.0L of pH 7.76 buffer using triethanolamine that is 0.020 M in both triethanolamine and triethanolammonium ion?

Answer 1

pKa of triethanolamine = 7.8

Using the Henderson-Hasselbach equation, we can find the ratio of moles between the conjugate base and the acid:

pH = pKa + log (triethanolammonium ion / triethanolamine)

7.76 = 7.8 + log (triethanolammonium ion / triethanolamine)

-0.04 = log (triethanolammonium ion / triethanolamine)

(triethanolammonium ion / triethanolamine) = 10^-0.04

(triethanolammonium ion / triethanolamine)= 0.91

triethanolammonium ion = 0.91 triethanolamine

we want to make a 0.020 M buffer, then you would have:

0.020 =(triethanolammonium ion + triethanolamine)

0.020 = 0.91 triethanolamine + triethanolamine

0.020 = 1 .91 triethanolamine

Triethanolamine = 0.010 moles

Molarity = number of moles / volume in L

Volume = 0.010 moles /1.0 M

= 0.01 L = 10 ml

triethanolammonium ion = 0.91 triethanolamine

= 0.91*0.010 moles

= 0.0091 moles

Molarity = number of moles / volume in L

Volume = 0.0091 moles /1.0 M

= 0.0091 L = 9 ml