Question

Develop 1.0 L of a buffer that maintains a pH of 6.30. Specify the acid-conjugate base pair or vise versa and the reasonable concentrations of other reagents needed (0.0100 M to 1.0 M). also discuss and limitations this buffer may have.

Answer 1

ph = 6.30

V = 1 Liter

acid /conjugate

pH = pKa + log(conjugate/acid)

nearest pKa --> H2CO3 and HCO3- data

6.30 = 6.37 + log(HCO3- / H2CO3)

10^(6.30 - 6.37) = (HCO3- / H2CO3)

(HCO3- / H2CO3) = 0.8511

Assume total molarity of buffer

M total = 0.1, then total moles MV = 0.1*1 = 0.1

HCO3- + H2CO3 = 0.1

(HCO3- / H2CO3) = 0.8511

HCO3- = 0.8511*H2CO3

substitute

0.8511*H2CO3 + H2CO3 = 0.1

H2CO3 = (0.1)/(0.8511+1) =

H2CO3 = 0.0540 mol

HCO3- = 0.8511*H2CO3 = 0.8511*0.0540 = 0.0459594 mol of conjguate base

total H2CO3 = 0.1

initially:

add [H2CO3] = 1 M stock solution --> 0.1 Liter or 100 mL

add 100 mL of H2CO3

then add

0.0459594 mol of NaOH, either via solid --> mass = mol*MW = 0.0459594 *40= 1.838376 g

or via voume

Assume stock NaOh --> 1 M

V = mol/M = 0.0459594 /1 = 0.0459594 L = 45.96 mL of NaOH

then, add total V to V = 1Liter

instructions

a= 100 mL of H2CO3

b= 1.838 g of NaOH OR 45.96 mL of a 1 M NaOh solution

c= add water (deiojnized) = up to V = 1000 mL or 1 L