Question

# ４０．2.20 L of a buffer solution that is 0.190 M in benzoic acid, C6H5COOH, and 0.340...

４０．2.20 L of a buffer solution that is 0.190 M in benzoic acid, C6H5COOH, and 0.340 M in sodium benzoate, C6H5CONa, are diluted to 4.40 L with distilled water. What is the pH of the buffer? Ka of C6H5COOH = 6.4 ✕ 10-5.

though you diluted pH doesnt change

because

pH =pKa + log(salt/Acid)

here concentrations are in the ratio what ever changes you done it will get cancelled in the ratio term

intial pH = 4.2 + log(0.34/0.19)

pH = 4.45

lets prove by doing all the calculations

intial

no of moles of C6H5COOH = molarity x Volume in liters = 0.19 M x 2.20 = 0.418 moles

no of moles of C6H5CONa = 0.34 x 2.2 = 0.748 moles

after dilution total volume will be 4.4L

concentration of C6H5COOH = 0.418 mol / 4.4 = 0.095M

concentration of C6H5CONa = = 0.748 / 4.4 = 0.17M

from Ka calculate pKa =

pKa = -log(Ka) = -log(6.4 ✕ 10-5) = 4.2

use the handerson equation

pH = pKa + log([C6H5CONa] / [C6H5COH]

pH = 4.2 + log(0.17/0.095)

pH = 4.2 + 0.25

pH = 4.45