A chemistry graduate student is given 500. mL of a 0.40 M benzoic acid HC 6 H 5 CO 2 solution. Benzoic acid is a weak acid with = K a × 6.3 10 ? 5 . What mass of NaC 6 H 5 CO 2 should the student dissolve in the HC 6 H 5 CO 2 solution to turn it into a buffer with pH = 4.45 ? You may assume that the volume of the solution doesn\'t change when the NaC 6 H 5 CO 2 is dissolved in it.
Solution
According to Henderson-Hasselbalch equation
pH = pKa + log { (salt) /( acid)}
pH = pKa + log (C6H5COONa/C6H5COOH)
pKa = -log(Ka)
pKa = - log ( 6.3×10^-5)
pKa = 4.20
Number of moles of C6H5COOH:
500 mL × ( 1L /1000 mL ) × ( 0.40 mol / L)
= 0.20 mol
Substitute all the values in above equation.
4.45 = 4.20 + log (C6H5COONa / 0.20 mol)
0.25 = log ( C6H5COONa / 0.20 mol)
C6H5COONa = 1.7783 × 0.2 mol
C6H5COONa = 0.35565 mol
Mass of C6H5COONa = 0.35565 mol × ( 144.11 g / 1 mol)
= 51.25 g
.
A chemistry graduate student is given 500- mL of a 0-40 M benzoic acid.docx
1. A chemistry graduate student is given 500. mL of a 0.40 M benzoic acid HC 6 H 5 CO 2
solution. Benzoic acid is a weak acid with = K a × 6.3 10 ? 5 . What mass of NaC 6 H 5 CO 2
should the student dissolve in the HC 6 H 5 CO 2 solution to turn it into a buffer with pH = 4.45
? You may assume that the volume of the solution doesn't change when the NaC 6 H 5 CO 2 is
dissolved in it.
Solution
According to Henderson-Hasselbalch equation
pH = pKa + log { (salt) /( acid)}
pH = pKa + log (C6H5COONa/C6H5COOH)
pKa = -log(Ka)
pKa = - log ( 6.3×10^-5)
pKa = 4.20
Number of moles of C6H5COOH:
500 mL × ( 1L /1000 mL ) × ( 0.40 mol / L)
= 0.20 mol
Substitute all the values in above equation.
4.45 = 4.20 + log (C6H5COONa / 0.20 mol)
0.25 = log ( C6H5COONa / 0.20 mol)
C6H5COONa = 1.7783 × 0.2 mol
C6H5COONa = 0.35565 mol
