Determination of Amount of Acid…by

Back Titration

Experiment 7

Experiment 7

Goal:

� Determine amount of acid that can be

neutralized by a commercial antacid

Method:

� React antacid tablet with excess stomach acid

(HCl)

� Perform a back titration using standardized

NaOH solution

Big Picture

1) Determine amount of acid

(HCl) in flask

2) Neutralize some HClwith

antacid tablet (CaCO3,Mg(OH)2)

3) Titrate remaining acid with

base (NaOH)

4) Find amount of acid

neutralized by tablet

?, n tabletdneutralize =

HCl usedinitialHCl V[HCl]n ⋅=,

NaOHNaOHedHCl titrat VMn ⋅=

titratedHClinitialHCl

tabletdneutralize

nn

n

,,

,

−

=

Acid-Base Definitions

Acids

� generate H+ in water

� H+ donors

� excess H+

Bases

� generate OH- in water

� H+ acceptors

� Excess OH-

Equilibrium in Water

14

253 101 −++

×==== o@

--

weq ]][OH [H]][OHO [HKK

Small K ≡ equilibrium favors reactants

Review: H+ and OH-

[H+] 1 × 100 to 1 × 10-14 in water

[OH-] 1 × 10-14 to 1 × 100 in water

][H

K ][OH w-

+=

14101 −+

×== ]][OH[HK-

w

][OH

K ][H

-

w=

+

Review: As [H+] rises, [OH-] falls

H+H+

H+

OH-OH-OH-

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Acidic Neutral Basic

][H -pH+

= log 0 to 14 in water

Relationships

H+ OH-

[H+] >[OH-]

Acidic

solution

Basic

solution

[H+] =[OH-] [H+] <[OH-]

Neutral

solution

H2O

Strong Acids (exp. 7)

100% dissociation / good H+ donor

equilibrium lies far to right

HCl → H+ + Cl-

HCl H+ Cl- HCl H+ Cl-

Rela

tive

# m

ole

s

Before

dissociation

After

dissociation

Weak Acids (exp. 13)

<100% dissociation / not-as-good H+ donor

equilibrium lies far to left

HCO3- H+ + CO3

2-

HCO3- H+ CO3

2-

Rela

tive

# m

ole

s

Before

dissociation

After

dissociation

HCO3- H+ CO3

2-

Acid Dissociation Constant

Relative strength of acid

+

+ H A HA -

[HA]

]][A[H K

-

a

+

=

Strong acids: Ka is huge

Weak acids: Ka is small

Indicators are weak acids

HIn H+ + In-

Overview Exp. 7

0) Standardize NaOH with KHP

1) Find moles of acid, nH+

HCl(aq) → H+(aq) + Cl-

(aq)

(H+ is H3O+)

HCl(aq)“stomach”

Overview Exp. 7

2) Neutralize some with antacid

Mg(OH)2 + 2 HCl →

Mg2+ + 2Cl- + 2H2O

CaCO3 + 2 HCl →

Ca2+ + 2Cl- + H2O + CO2

Antacid

Mg(OH)2 + CaCO3

HCl(aq)

Overview Exp. 7

3) Back-titrate leftover HCl

with NaOH

HCl + NaOH →

Na+ + Cl- + H2O

NaOH

titrant

Leftover

HCl(aq)

Experimental Stages

used HClinitial V]HCl[n ⋅=

nfinal

ninitial

HCl consumed

by antacid

HCl consumed

by titration0n final =

n HClneutralized by tablet = n HClinitial ─ n HCltitrated

Part 1 Standardization of NaOH

Find molarity of NaOH

� g KHP(calculated in prelab) mKHP

in ~50 mL H2O + drops BTB

� Titrate with NaOH VNaOH

Given: ~0.5 M [NaOH]

@ eq. pt.: nNaOH = nKHP MM KHP=204.23g/mol

Calculate: nNaOH nNaOH

� 3 trials

� Calculate average MNaOH ⟨M⟩NaOH

Part 2 Standardization of HCl

Find HCl molarity

� Exactly 10mL HCl + drops BTB VHCl

� Titrate with NaOH VNaOH

Given: [NaOH] MNaOH

@ eq. pt.: nHCl = nNaOH

Find: nNaOH used = [NaOH] . VNaOH

Calculate: nHCl titrated in 10 mL MHCl

� Repeat

� Calculate average MHCl ⟨M⟩NaOH

Example data

Use MHCl in part 3 (titration)

Standardization Part 1 NaOH

Trial g KHP mL NaOH M NaOH

1 3.0000 29.4 0.4996

2 3.0000 29.1 0.5048

3 3.0000 30.9 0.4754

averages 3.0000 29.8 0.4929

Standardization Part 2 HCl

Trial mL HCl mL NaOH M HCl

1 10.00 20.5 1.0105

2 10.00 20.5 1.0105

averages 10.00 20.5 1.0105

Use MNaOH in parts 2 (HCl standardization) and 3 (titration)

Part 3

Antacid tablet reaction

� Rinse all glassware

� Record mass of 4 antacid tablets

� Label 4 125 mL Erlenmeyer flasks

� Add exactly 25 mL HCl and 1 tablet to each

� Gently boil 5 minutes

� Cool and add 2 drops BTB indicator

� First trial = guide for 3 more

� Titrate with NaOH to equivalence point

� Save as standard color

� Repeat more 3 times

Why is heating necessary?

Heat removes CO2 made in HCl/antacid reaction

CaCO3 + 2 HCl CaCl2(aq) + H2O + CO2

CO2 + H2O H2CO3(aq) HCO3- + H+

Removing CO2 allows maximum amount of acid to be

neutralized

Part 3 Acid Neutralized by Tablet

�NaOH + HClremaining → NaCl(aq) + H2O

Given: [NaOH], [HCl] MNaOH, MHCl

Find: nNaOH used VNaOH

@ eq. pt.: nNaOH = nHCl

Calculate: nHCl remaining nHCl

�nHCl titrated = nNaOH used = [NaOH] . VNaOH

Example dataTitration Part 3 HCl

Trial g tablet mL HCl mL NaOH mol HCl titrated1 1.50 25.00 22.2 0.0143

2 1.51 25.00 21.4 0.0147

3 1.47 25.00 22.0 0.0144

4 1.50 25.00 21.5 0.0147

averages 1.50 25.00 21.8 0.0145

avg mol neutralized/g tablet 0.00972

Theoretical amount neutralized by tablet

g base MM mol ratio mol HClneutralizedMg(OH)2 0.11 58.33 2 0.0038

CaCO3 0.55 100.08 2 0.0110

sum 0.0148

%difference 1.58

Part 3 nHCl neutralized by tablet

HCl usedinitial V[HCl]n ⋅=

nfinal

ninitial

HCl consumed

by antacid

HCl consumed

by titration

n HClneutralized by tablet = n HClinitial ─ n HCltitrated

NaOHNaOHedHCl titrat VMn ⋅=

Part 3 Reporting ResultsAcid neutralized by tablet

110 mg Mg(OH)2 + 2 HCl CaCl2(aq) + H2O + CO2

550 mg CaCO3 + 2 HCl CaCl2(aq) + H2O + CO2

1.) Per mass

average, σ

2.) % acid neutralized relative to predicted

% of moles

actually neutralized

vs. predictedstoichimetry

tablet

alexperiment HCl

g

n3 good trials

%100n

n

al theoreticHCl

alexperiment HCl×

calculated

average

Report

Abstract

Results

Sample calculations including

� Part 1: individual [NaOH] + average & σ

� Part 2: individual [HCl] + average & σ

� Part 3: moles HCl initially

moles HCl titrated with NaOH

moles HCl neutralized by tablet

moles HCl per mass tablet

% predicted

Discussion/review questions

Strong acid-strong base titrations

0

1.0

Vol NaOH added (mL)

50.0

7.0

13.0

pH

100.0

Equivalencepoint

At equivalence point:

mol H+ = mol OH-

nH+ present= nOH- added