[PDF] Worksheet 22 – Weak Acid/Strong Base Titrations A. Initial pH This

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Worksheet 22 - Weak Acid/Strong Base Titrations

A. Initial pH

This is determined by the initial concentration of the weak acid and the dissociation constant, K a.

HAAOHK

a 3 Set up an ICE table for the dissociation of the weak acid in water. The "change" line will include entries like "-x" and "+x" because the weak acid will only partially dissociate. Solve for [H 3 O ] and use it to calculate the pH: pH = -log[H 3 O Be aware that polyprotic acids behave a little differently than monoprotic acids and will be discussed next week.

B. pH during the titration

For every mole of OH

added to a weak acid, one mole of acid will be neutralized. Neutralization of 1 mole of a weak acid (HA) will produce one mole of water and one mole of its conjugate base (A

HA + OH

H 2 O + A Since significant amounts of both the weak acid AND its conjugate base are present, the pH can be calculated using the Henderson-Hasselbalch equation: acidbasepKpH a log This can only be used after the initial point and before the equivalence point, when the acid and its conjugate base are both major species in the solution. At the half-way point, when exactly one half of the HA has been converted to A by addition of OH , pH = pK a , because log (1) = 0.

C. Equivalence point

The equivalence point is reached when equal moles of acid and base are mixed. At the equivalence point, all of the HA will be converted to A , the conjugate base of the initial weak acid. The conjugate base (and water) will be the major species in the solution. There will be as many moles of these conjugate species as there were moles of acid or base at the beginning of the titration. The volume will be the sum of the initial volume of weak acid and the volume of base added to bring the solution to the equivalence point. In the case of a weak acid titrated with a strong base, A + H 2

O HA + OH

The extent of this reaction will be determined by the K b of the conjugate base, A This can be calculated from the following relationship: baw KKK 14 100.1
The resulting solution will be basic. The pH is calculated by setting up an ICE table for the dissociation of the conjugate base in water, and using [OH ] to solve for pOH.

D. Above the equivalence point

Above the equivalence point, any added strong base will determine the pH. The excess OH divided by the volume can be used to determine the pOH (and therefore pH) of the solution.

1. Consider the titration of 50.0 mL of HNO

2 (K a = 4.0x10 -4 ) with KOH.

At each step of the titration...

a) write a reaction to show the initial reaction upon mixing b) construct an ICE table to represent the reaction (Should you use concentration or moles?) c) determine the major species present after the reaction is complete d) calculate the pH of the solution Complete these steps for the addition of 0.0 mL, 25.0 mL, 50.0 mL, 90.0 mL,

100.0 mL, 110.0 mL and 150.0 mL of KOH.

0.0 mL KOH

a) HNO 2 + H 2 O NO 2- + H 3 O b) [HNO 2 ] [NO 2- ] [H 3 O

Initial 2.0 0 0

Change -x +x +x

Equil. 2.0 - x x x

This ICE table uses units of concentration because the volume is fixed c) The major species present at equilibrium are HNO 2 and H 2

O (weak acid

solution) d) pH is calculated using the concentration of [H 3 O

55.11083.2log valid)is rule (5% 1083.20.20.2100.4

2222
2 324
pHxx xx

HNOOHNOK

a

25.0 mL KOH

a) HNO 2 + OH NO 2- + H 2 O b) This ICE table uses units of moles because the volume changes when the

KOH is added

22
2

HNO mol 100.0L1HNO mol 2.0L 050.0HNO mol

OH mol 025.0NaOHmol1.0OH mol 1.0

L 1NaOH mol 1.0L 025.0OH mol

moles HNO 2 OH NO 2-

Initial 0.100 0.025 0

Change -0.025 -0.025 +0.025

Equil. 0.075 0 0.025

c) The major species present at equilibrium are HNO 2 , NO 2- , K and H 2

O (buffer

solution) d) pH is calculated using the Henderson-Hasselbalch equation

40.3100.4loglog

4 aa KpK

92.2075.0025.0log40.3acidbaselog

a pKpH

50.0 mL KOH

a) HNO 2 + OH NO 2- + H 2 O b) This ICE table uses units of moles because the volume changes when the

KOH is added

OH mol 050.0NaOHmol1.0OH mol 1.0

L 1NaOH mol 1.0L 050.0OH mol

moles HNO 2 OH NO 2-

Initial 0.100 0.050 0

Change -0.050 -0.050 +0.050

Equil. 0.050 0 0.050

c) The major species present at equilibrium are HNO 2 , NO 2- , K and H 2

O (buffer

solution) d) pH is calculated using the Henderson-Hasselbalch equation

40.3050.0050.0log40.3acidbaselog

a pKpH **This is the halfway point, so pH = pK a

90.0 mL KOH

a) HNO 2 + OH NO 2- + H 2 O b) This ICE table uses units of moles because the volume changes when the

KOH is added

OH mol 090.0NaOHmol1.0OH mol 1.0

L 1NaOH mol 1.0L 090.0OH mol

moles HNO 2 OH NO 2-

Initial 0.100 0.090 0

Change -0.090 -0.090 +0.090

Equil. 0.010 0 0.090

c) The major species present at equilibrium are HNO 2 , NO 2- , K and H 2

O (buffer

solution) d) pH is calculated using the Henderson-Hasselbalch equation

35.4010.0090.0log40.3acidbaselog

a pKpH

100.0 mL KOH

a) HNO 2 + OH NO 2- + H 2 O b) This ICE table uses units of moles because the volume changes when the

KOH is added

OH mol 100.0NaOHmol1.0OH mol 1.0

L 1NaOH mol 1.0L 100.0OH mol

moles HNO 2 OH NO 2-

Initial 0.100 0.100 0

Change -0.100 -0.100 +0.100

Equil. 0 0 0.100

c) The major species present at equilibrium are NO 2- , K and H 2

O (weak base

solution) **This is the equivalence point of the titration because the initial mole values of weak acid and strong base were equivalent d) pH is calculated using the K b for the weak base; set up a new ICE table using concentration units (the volume of this solution is fixed now) NO 2- + H 2 O OH + HNO 2 22
2

NO M 667.0L100.0L 050.0NO mol 0.100NO

[NO 2- ] [OH ] [HNO 2

Initial 0.667 0 0

Change -x +x +x

Equil. 0.667-x x x

61.81439.51008.4log valid)is rule (5% 1008.4667.0667.0105.2100.4100.1

6622
2 2 11 414
pOHpHpOHxx xx

NOOHHNOK

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