Question

CHEMISTRY WORKSHEET Name CHEM 1120 RECITATION Date BUFFERS 1. Your employer wants you to make a 500.0 ml buffer that contains 1.000 M benzoic acid and has a pH of 4.30. If you choose sodium benzoate to be the base, how many grams of benzoic acid and how many grams of sodium benzoate must you dissolve in water to create your buffer? Ke of Benzoic acid = 6.4 x 10-5 Benzoic acid = C H8O2 Sodium benzoate = NaC.HsO2 What will be the pH of this buffer after the addition of 25.00 mL of 1.000 M HCI?

Answer 1

1. The buffer solution is prepared by mixing benzoic acid and sodium benzoate.

Use the Handerson-Hasselbalch equation to determine the number of moles of the salt needed to produce the buffer solution with the desired pH.

Ka of benzoic acid= 6.4*10-5

therefore pKa= -log(Ka) = -[ log(6.4*10-5)]= 4.20

The pKa value for benzoic acid is 4.20.

Moles of acid can be calulated by

nAcid=CM×V

nAcid=1.0 mol.L×0.500L

nAcid= 0.50

−−−−−−−−−−−−−−−−

Solve for the unknown mole of salt (nSalt), by using Henderson-Hasselblch equation

pH=pKa+log(nSalt/nAcid)

4.30=4.20+log(nSalt/nAcid)

log(nSalt/nAcid)=4.30−4.20

log(nSalt/nAcid)=0.10

nSalt/nAcid=10(0.10)

nSalt/nAcid=1.26

nSalt=0.50×1.1

nSalt=0.55 mol.

Therefore gram of sodium benzoate mSalt= mole* molecular wt.= 0.55mol.×144.11gmol.

mSalt≅79.26 gm

similarly gram of sodium benzoate mAcid= 122.12*0.55=67.166gm

2.

Benzoic acid dissociates:

HA(aq)⇌H+(aq)+A− (salt)

The no. H+ added is given by:

nH+= Conc× Volume= 1×25/1000=0.025

These H+ ions will be absorbed by the large reserve of salt ions shifting the equilibrium thus "buffering" the pH so that the no. of moles of salt remaining is given by:

nSalt−=0.55−0.025=0.525

Each mole of H+ added will form 1 mole of HA so no. moles HA formed =0.025

hence total moles of acid

nAcid= 0.50+0.025= 0.525

Hence

pH=pKa+log(nSalt/nAcid)

=4.20+log(0.525/0.525)

pH=4.20