http://lawrencekok.blogspot.com

Prepared by Lawrence Kok

IB Chemistry Titration Techniques, Acid/Base Titration and IA on Titrations.

Volumetric pipette

Indicator

Retort stand

• Quantitative analysis - determine unknown conc of an analyte

• Add known conc (std) from burette to unknown conc in flask.

• Titrant/titrator (added) + analyte/titrand (to be analyzed)

Conical flask

Pipette filler

Molarity- 0.1M Na2CO3

Titration

Titration Set up

BuretteTitrant/Titrator

Conical flaskAnalyte/Titrand

Notes/ sample titration calculation

White tile

Volumetric flask

Standardization of (ACID) with standard (BASE)

10.6g Na2CO3 10.6g in 1 L

Burette

Preparation of std (BASE) Preparation of std (ACID)

Titration

Redox Titration Acid Base Titration Complexometric titration

Neutralization

IndicatorpH sensor Conductometric

Changes colour (end point) at equivalent point

Indicator pKa pH rang

e

Colour Acid

Colour Base

Methyl orange 3.46 3.2- 4.4

RED Yellow

Bromophenol Blue

4.10 3.0- 4.6

Yellow Blue

Bromocresol Green

4.90 3.8- 5.4

Yellow Blue

Methyl Red 5.00 4.8- 6.0

Red Yellow

Bromothymol Blue

7.30 6.0- 7.6

Yellow Blue

Phenol Red 8.00 6.6- 8.2

Yellow Red

Phenolphthalein 9.50 8.2- 10.0

Colourless

Pink

Video on pH sensor Video on indicator Video on conductometric

measured usingmeasured using

Video on conductometric

3

1

62

4 5

Titration Steps

Rinse burette – titrant

Bottom fill with titrant NOT air gap

Fix it on retort stand read (Bottom of meniscus)

Record vol to nearest 0.01 ml9

8 7

Final vol to nearest 0.01 ml

Click here to view video

Rinse burette – deionized water Add dropwise – end point is near

Titration

Redox Titration Acid Base Titration Complexometric titration

Neutralization

Condition for Acid/Alkali TitrationOne reactant – must be std (known conc) or capable being standardised

Equivalent point – equal amt neutralize each otherEnd point detectable by colour change, pH change /conductivity

Acid/Base as primary standard -Stable/solid

-Soluble in water-Does not decompose over time

Primary std acid- Potassium hydrogen phthalate

Primary std base- Anhydrous sodium carbonate

Standard 0.1M Na2CO3

10.6g in 1 L

Volumetric Burette

Unable to prepare accurate conc of NaOH/HCI

•hygroscopic nature NaOH – absorb water vapour

• HCI is in vapour state – diff to measure amt

VolumetricBurette

Standard 0.1M KHP

20.4 g KHP

20.4 g in 1L

Unknown Conc NaOH

Unknown Conc HCI

? ?

Standardize NaOH

using KHP

Standardize HCI

using Na2 CO

3

10.6 g Na2CO3

0.1M KHP – 0.1 mole of KHP in total vol of solution (1L)

Primary standard acid

Mass of KHP → 0.1 mole KHP x M = 0.1 x 204.22gStep 1

Step 2

Pour to 1L volumetric flask Step 3

Add water until 1L mark

Transfer to beaker, add water to dissolve

Step 4

Step 5

20.4 g

Video std solution preparation

Molarity = 0. 1 mole (0.1M) 1 L total vol (solute + solvent)

Primary standard

Preparing std solution – 0.1 M – 0.1 mole KHP in 1 L

0.1M Na2CO3 – 0.1 mole of Na2CO3 in total vol of solution (1L)

Primary standard base

Mass of Na2CO3 → 0.1 mole Na2CO3 x M = 0.1 x 106 gStep 1

Step 2

Pour to 1L volumetric flask Step 3

Add water until 1L mark

Transfer to beaker, add water to dissolve

Step 4

Step 5

10.6 g

Video std solution preparation

Molarity = 0. 1 mole (0.1M) 1 L total vol (solute + solvent)

Primary standard

Preparing std solution – 0.1 M – 0.1 mole Na2CO3 in 1 L

Pipette 25 ml NaOH in conical flask

Step 2

Fill burette with std 0.1M KHP sol

Step 3

Step 4

Step 5

2 drop phenolthalein – colourless to pink

Ini vol KHP recorded

Titrate until pink colour fades away

Final vol KHP recordedStep 6

Step 7 Repeat till consistent result agree within 0.1 cm3 (triplicate)

Standardization (ACID) with standard (BASE)

? Conc NaOH

Standardization (BASE) with std (ACID)

Step 1 Step 2

Step 3 Step 4

Step 5 Step 6 Step 7

Fill burette with std 0.1M Na2CO3 sol

Ini vol Na2CO3 recorded

Pipette 25 ml HCI in conical flask

2 drop methyl orange to HCI – red

Titrate until red changes to pink

Final vol Na2CO3recorded

Repeat till consistent result agree within 0.1 cm3 (triplicate)

Std 0.1M Na2CO3

? Conc HCI

Step 1 Std 0.1M KHP

Standardization of (BASE) with std (ACID)

Vol KHPFin vol = 29.50 ± 0.05Ini vol = 3.10 ± 0.05 Vol = 26.4 ± 0.1

Uncertainty in vol Add abs ∆ for final + ini= (0.05 + 0.05) = ± 0.1

Ave Vol KHP ± uncertainty= 26.4 + 26.4 + 26.4 3= (26.4± 0.1) cm3

Conc NaOH = 0.106 ± 0.5%

0.5 x 0.106 100= 0.001

Conc NaOH = 0.106 ± 0.001M

Lit value - NaOH = 0.100MExpt value– NaOH = 0.106MDifference = 0.006% Error – Difference x 100% Lit value 0.006 x 100 % = 6% 0.100

Data Processing Data collection

Vol KHP /cm3Titration trials

1 2 3

Final vol/(±0.05)cm3

38.50 29.50 45.90

Ini vol/ (±0.05)cm3

12.10 3.10 19.50

Total vol/ (±0.1)cm 3

26.4 26.4 26.4

NaOH + KHP → NaKP + H2O M = ? M = 0.1V = 25ml V = 26.4 ml

MMVMVM

b

aa

bb

106.011

0264.01.0025.011

Error Analysis

% ∆ - burette % ∆ - pipette

Abs ∆ vol x 100% Ave vol = 0.1 x 100% 26.4 = 0.38%

Abs ∆ vol x 100% vol = 0.03 x 100% 25.00 = 0.12 %

Total % ∆ = % ∆ burette + %∆ pipette = 0.38% + 0.12% = 0.5%

Data Processing

% uncertainty(Abs uncertainty)

Acid (KHP)

Base (NaOH)

Standardization of (ACID) with std (BASE)

Vol Na2CO3

Fin vol = 29.50 ± 0.05Ini vol = 3.10 ± 0.05 Vol = 26.4 ± 0.1

Uncertainty in vol Add abs ∆ for final + ini= (0.05 + 0.05) = ± 0.1

Ave Vol Na2CO3 ± uncertainty= 26.4 + 26.4 + 26.4 3= (26.4 ± 0.1) cm3

Conc HCI = 0.2112 ± 0.5%

0.5 x 0.2112 100= 0.001

Conc HCI = 0.2112 ± 0.001M

Lit value - HCI = 0.200MExpt value – HCI = 0.2112MDifference = 0.011% Error – Difference x 100% Lit value 0.011 x 100 % = 6% 0.200

Data Processing Data collection

Vol Na2CO3 /cm3Titration trials

1 2 3

Final vol/(±0.05)cm3

38.50 29.50 45.90

Ini vol/ (±0.05)cm3

12.10 3.10 19.50

Total vol/ (±0.1)cm 3

26.4 26.4 26.4

Na2CO3 + 2HCI → 2NaCI + H2O + CO2 M = 0.1 M = ?V = 26.4 ml V = 25.0 ml

MM

VMVM

a

aa

bb

2112.021

0250.00264.01.021

Error Analysis

% ∆ - burette % ∆ - pipette

Abs ∆ vol x 100% Ave vol = 0.1 x 100% 26.4 = 0.38%

Abs ∆ vol x 100% vol = 0.03 x 100% 25.00 = 0.12 %

Total % ∆ = % ∆ burette + %∆ pipette = 0.38% + 0.12% = 0.5%

Data Processing

% uncertainty(Abs uncertainty)

Base (Na2CO3)

Acid (HCI)

NaOHM = ? V = 25.0ml

KHPM = 0.100MV = 26.4 ml

HCI M = ? V = 25.0ml

Sample Titration Calculation

Na2CO3 M = 0.100MV = 26.4 ml

Standardization of (BASE) with std (ACID)

Standardization of (ACID) with std (BASE)

KHP + NaOH → NaKP + H2O M = 0.100 M = ?V = 26.40 ml V = 25.0 ml

MM

VMVM

b

bb

aa

106.011

0250.00264.01.011

Na2CO3 + 2HCI → 2NaCI + H2O + CO2 M = 0.100 M = ?V = 26.4 ml V = 25.0 ml

MM

VMVM

a

aa

bb

2112.021

0250.00264.01.021

Click here titration simulation Click here titration simulation Click here titration simulation Click here titration simulation

Simulation on Titration

Sample Titration Calculation

MMVMVM

b

aa

bb

12.212

5.026.00.10250.012

32012

305.05.1

12

cmVVMVM

b

aa

bb

25 ml NaOH require 26.5cm3 of 1.0M H2SO4

for neutralization.Find its molarity of NaOH.

Find vol of 1.5M NH3 required to neutralize

30 ml of 0.5M H2SO4

H2SO4M = 1 MV = 26.5 ml

NaOHM = ? V = 25 ml

2NaOH + H2SO4 → Na2SO4 + 2H2OM = ? M = 1MV = 25.0ml V = 26.5ml

H2SO4M = 0.5MV = 30ml

NH4OHM = 1.5MV = ? ml

2NH4OH + H2SO4 → (NH4)2SO4 + 2H2O M = 1.5M M = 0.5MV = ? ml V = 30.0ml

Simulation on Titration

Click here titration simulation Click here titration simulation Click here titration simulation Click here titration simulation

Sample Titration Calculation

32521

0.2505.021

cmV

VMVM

a

aa

bb

MM

VMVM

a

aa

bb

16.021

25102.021

Simulation on Titration

Click here titration simulation Click here titration simulation Click here titration simulation Click here titration simulation

Find vol of 2.0M HCI needed to neutralize 0.5M Na2CO3 in 50ml water.

HCIM = 2.0MV = ? ml

Na2CO3M = 0.5M

V = 50ml

Na2CO3 + 2HCI → 2NaCI + CO2 + H2OM = 0.5M M = 2.0MV = 50ml V = ? ml

Na2CO3 M = 0.2MV = 10.0ml

HCI M = ? V = 25 ml

10 cm3 of 0.2 M Na2CO3 need 25 cm3 of HCI for neutralization. Find molarity of

HCI.

Na2CO3 + 2HCI → 2NaCI + H2O + CO2M = 0.2M M = ?V = 10 ml V = 25 ml

Titration for IA assessment

Acid Base Titration

Standardization HCI with std Na2CO3

Standardization NaOH with std KHP

Titration bet NaOH with std HCI

Titration bet HCI

with std NaOH

Determine water crystallization in hydrated Na2CO3 with std HCI

Standardization KMnO4 with std ammonium iron(II) sulphate

Fe 2+ in iron pillwith std KMnO4

Hypochlorite (OCI-) in bleach with iodine/thiosulphate

Determine ethanoic acid in vinegar

Cu 2+ in brass with iodine/thiosulphate

Standardization KI/I2 with std KIO3

Determine acetylsalicylic acid in aspirin

Vit C in fruits with iodine/thiosulphate

Standardization Expt Acid/Base Expt

Standardization Expt Redox Expt

Redox Titration

Standardization KI/I2 with std sodium thiosulphate

Iodine/thiosulphate (iodometric titration)

Equilibrium established when ethanoic acid and ethanol react together

in strong acid, using propanone as solvent. Eqn given.

CH3COOH + C2H5OH ↔ CH3COOC2H5 + H2O Density ethanoic acid is 1.05 g cm–3.

i.Find amt, mol, of acid presentii.Conc acid is 1.748 mol dm–3. Find % uncertainty of conc.

Titration performed on acid using a base. Result shown below

Find absolute uncertainty of titre for Titration 1 (27.60 cm3).

Liquid Vol/cm3 Ethanoic acid 5.00 ±

0.05

Ethanol 5.00 ± 0.05

Hydrochloric acid 1.00 ± 0.02

Propanone 39.0 ± 0.5

gmassvoldenstiymass

25.500.505.1

volmassDensity

molMol

RMMmassMol

0874.06025.5

RMM acid = 60

% uncertainty acid conc = % uncertainty in vol acid + % uncertainty in total vol

volmolacidConc .

(0.05/5.00) x 100 % = 1 %

(0.62/50) x 100% = 1.24 %

Total % uncertainty = (1 + 1.24) % = 2.24%

Uncertainty final – initial vol (28.80 ±0.05 – 1.20 ±0.05 ) = (27.60 ± 0.1)

Add absolute uncertainty together

Two rxn kinetic investigated using iodine clock rxn.

Reaction A: H2O2 + 2I− + 2H+ → I2 + 2H2 OReaction B: I2 + 2S2O3

2− → 2I− + S4O62-

i. Find total uncertainty, in vol of rxn mixtureMixture contained:

5.0 ± 0.1 cm3 of 2M H2O2

5.0 ± 0.1 cm3 of 1 % starch20.0 ± 0.1 cm3 of 1M H2SO4

20.0 ± 0.1 cm3 of 0.01 M Na2S2O3

50.0 ± 0.1 cm3 of water with 0.0200 ± 0.0001 g KI

i. Add all vol together: Add all absolute uncertainty together.(5.0 ± 0.1) + (5.0 ± 0.1) + (20.0 ± 0.1) + (20.0 ± 0.1 ) + (50.0 ± 0.1) = (100 ± 0.5) cm 3

ii. Conc KI =Mass/ vol% uncertainty conc KI = % uncertainty mass + % uncertainty vol KI% ∆ mass = (0.0001/0.02) x 100% = 0.5 %% ∆ vol = (0.1/50) x 100% = 0.2 %% conc KI = (0.5 + 0.2)% = 0.7 %

iii. Final Conc KI = Conc KI in total mixture

ii. Find % uncertainty for KI conc in final rxn sol.

iii. Find % uncertainty for KI conc in overall rxn mixture

% ∆ final conc KI = % ∆ conc KI + % ∆ total vol KI

% ∆ conc KI = 0.7 %

% ∆ total vol = (0.5/100) x 100 % = 0.5% % conc KI

= (0.5 + 0.7) = 1.2 %

Mixture contained:5.0 ± 0.1 cm3 of 2M H2O2

5.0 ± 0.1 cm3 of 1 % starch20.0 ± 0.1 cm3 of 1M H2SO4

20.0 ± 0.1 cm3 of 0.01 M Na2S2O3

50.0 ± 0.1 cm3 of water with 0.0200 ± 0.0001 g KItotal vol

Only vol/mass/conc KI

4.32 x 10-5 x 176.14 = 7.61 x10-3 g Vit C

KIO3 + 5KI + 6H+ → 3I2 + 6K+ + 3H2O

3C6H8O6 + 3I2 → 3C6H6O6 + 6I- +

6H+

Iodometric titration on Vit C, (C6H8O6). Vit C titrated with 0.002M KIO3 , using excess KI

and starch.

Redox Titration – Vit C quantification

KIO3

M = 0.002M

Vit CAmt = ?

Mole ratio (1 :3)1 mol KIO3 : 3 mol I2 : 3 mol C6H8O6

1 mol KIO3 3 mol C6H8O6

Amt = ?

transfer1g KI excess + starch

titrated

Vit C

5

3

1032.4)..(

31

)..(0072.0002.0

31

).()(

CVitMole

CVitMole

CVitMVKIOMV

i. Find mass, of KIO3, required to prepare 0.250 dm3 of 0.002M KIO3

ii Titration results shown in table belowFind % uncertainty in mean vol of KIO3 used.

Mean vol = (7.20 ± 0.10) cm3

Find amt of KIO3 used

Mol = M x V = 0.002 x 7.20 1000 = 1.44 x 10-5 mol

41000.5

250.0002.0250.0

002.0

.

mol

mol

molvolmolacidConc

Convert mole KIO3 → Mass/g

X RMM = 214.00

5.00 x 10-4 x 214.00 = 0.107 g

% ∆ vol = (0.10/7.20) x 100 % = 1.4 %Find amt, Vit C in

sampleFind mass of Vit CConvert mole Vit C → Mass

RMM Vit C – 176.14

M x 0.0292 = 2.5 x 10-3

acid M = 2.5 x 10-3

0.0292 M = 0.0856M

Acid/Base Titration– Ethanoic acid in vinegar

CH3COOH M = ? V = 29.2ml

NaOH M = 0.1MV = 25.0ml

NaOH + CH3COOH → CH3COONa + H2O M = 0.1M M = ?V = 25ml V = 29.2ml

V = 250ml M = ?

Mole ratio (1 : 1) 1 mole NaOH - 1 mole acid 2.5 x 10-3 mole NaOH - 2.5 x 10-3 acid

Mole ratio – 1: 1

Diluted 10x

V = 25 ml M = ?

25ml of conc vinegar (ethanoic acid) was diluted to total vol 250 ml in a flask. Diluted vinegar was transfer to a burette. 25ml, 0.1M NaOH is pipette into a flask, with indicator added. End

point reached when average 29.2 ml of diluted vinegar added. Find its molarity.

mole ratio

Moles bef dilution = Moles aft dilution M1 V1 = M2V2M1 = Ini molarity M2= Final molarityV1 = Ini vol V2 = Final vol

Mole NaOH = MV = (0.1 x 0.025) = 2.5 x 10-3

0856.011

0292.0025.01.011

a

aa

bb

M

VMVM

formula

MMM

VMVM

856.02500856.025

1

1

2211

Acid/Base Titration - Empirical formula Na2CO3. x H2O

HCI M = 0.100 M V = 48.8ml

Na2CO3 M = ? MV = 25 ml

2HCI + Na2CO3 → 2NaCI + CO2 + H2O M = 0.1M M = ?V = 48.8ml V = 25.0ml

V = 1L M = ?

25 ml transfer

Mole ratio – 2: 1

Mass Na2CO3 . x H2O = 27.82 gMass Na2CO3 = 10.36 gMass of water = (27.82 – 10.36) g = 17.46 g

Diuted to 1L

27.82gNa2CO3. xH2O

27.82 g of hydrated (Na2CO3 . x H2O) dissolved in water, making up to 1L. 25 ml of sol was neutralized by 48.8ml of 0.1 M HCI. Cal molarity and mass of Na2CO3 present in 1L of sol. Find x

Convert mol dm-3 → g dm-3 Empirical formula

Na2CO3 H2O

Mass/g

10.36 17.46

RMM 106 18.02

Mole 10.34/106= 0.09773

17.46/18.02= 0.9689

Lowest

ratio

0.09773/0.09733 1

0.9689/0.09733 10

Empirical formulaNa2CO3 . 10 H2O

MM

VMVM

b

bb

aa

0976.012

0250.00488.01.012

0.0976 x 106 = 10.36g/dm3

X RMM

Redox Titration - % Fe in iron tablet

Iron tablet contain FeSO4.7H2O. One tablet weigh 1.863 g crushed, dissolved in water and sol made up to 250 ml. 10ml of this sol added to

20 ml of H2SO4 and titrated with 0.002M KMnO4. Ave 24.5ml need to reach end point. Cal % Fe in iron tablet.

10ml transfer 20ml acid added

1.863 g250ml

KMnO4

M = 0.002MV = 24.5 ml

Fe2+

M = ? V = 30ml

MnO4- + 5Fe2+ + 8H+ → Mn2+ +

5Fe2+ + 4H2OM = 0.002M M = ? V = 24.5ml

Mole ratio – 1: 5

Mass (expt yield) = 1.703gMass (Actual ) = 1.863g% Fe = 1.703 x 100% 1.863 = 91.4%

6.125 x 10-3 x 278.05 = 1.703 g FeSO4

10ml sol contain - 2.45 x 10-4 Fe2+

250ml sol contain - 250 x 2.45 x 10-4

Fe2+

10 = 6.125 x 10-3 mole Fe2+

42

2

1045.2.51

.0245.0002.051

FeMoleFeMole

VMVM

bb

aa

Convert mole → Mass

X RMM

Mole bef dil = Mole aft dil M1 V1 = M2V2

M1 x 10 = 1.78 x 10-2 x 250 M1 = 1.78 x 10-2

x 250 10 M1 = 0.445M

2CIO- + 2I- + 2H+ → I2 + 2CI- + H2O

I2 + 2S2O32- → S4O6

2- + 2I-

10ml bleach (CIO-) diluted to a vol of 250 ml. 20ml is added to 1g of KI (excess) and iodine produced is titrated with 0.0206 M Na2S2O3.Using

starch indicator, end point was 17.3ml. Cal molarity of CIO in bleach.

Redox Titration – CIO- in Bleach

Na2S2O3

M = 0.0206MV = 17.3ml

I2

M = ?

Mole ratio ( 1 : 1)2 mole CIO- : 1 mole I2 : 2 mole S2O3

2-

2 mole CIO- 2 mole S2O3

2-

10.0ml CIO-

transfer

V = 250mlM = 1.78 x 10-2 M

20ml transfer

1g KI excess added

M x V = Mol CIO- M x V = 3.56 x 10-4 M x 0.02 = 3.56 x 10-4 M = 3.56 x 10-4 002 M = 1.78 x 10-2 M diluted 25x

Diuted 25x

V = 10M = ?

titrated

Water added till 250ml

4

32

1056.3)..(22

0173.00206.0).(22

)()(

CIOMole

CIOMoleOSMV

CIOMV

KIO3 + 5KI + 6H+ → 3I2 + 6K+ + 3H2O

3C6H8O6 + 3I2 → 3C6H6O6 + 6I- +

6H+

Iodometric titration on Vit C, (C6H8O6). 25 ml Vit C titrated with 0.002M KIO3 from burette,

using excess KI and starch. Ave vol KIO3 is 25.5ml. Cal molarity of Vit C.

Redox Titration – Vit C quantification

KIO3

M = 0.002MV = 25.5ml

Vit CM = ?V = 25ml

Mole ratio (1 :3)1 mol KIO3 : 3 mol I2 : 3 mol C6H8O61 mol KIO3 3 mol C6H8O6

V = 25mlM = ?

25ml transfer1g KI excess + starch

titrated

Vit C

4

3

1053.1)..(

31

)..(0255.0002.0

31

).()(

CVitMole

CVitMole

CVitMVKIOMV M x V = Mol Vit C

M x V = 1.53 x 10-4 M x 0.025 = 3.56 x 10-4 M = 3.56 x 10-4 0025 M = 6.12 x 10-3 M

2.82 x 10-3 x 63.5 = 0.179 g Cu in 25ml

1.79 g Cu in 250ml

% Cu = mass Cu x 100% mass brass = 1.79 x 100% 2.5 = 71.8%

2Cu2+ + 4I- → I2 + 2CuI

I2 + 2S2O32- → S4O6

2- + 2I-

2.5g brass react with 10ml HNO3 producing Cu2+ ion. Sol made up to 250ml using water. Pipette 25ml of sol to flask. Na2CO3 add to neutralize excess acid. 1g KI (excess) and starch added. Titrate with 0.1M S2O3

2- and end point, is 28.2 ml. Find molarity Cu 2+ and % Cu found in brass.

Redox Titration - % Cu in Brass

Na2S2O3

M = 0.1MV = 28.2ml

I2M = ?

Mole ratio (1 : 1)2 mol Cu2+ : 1 mol I2 : 2 mol S2O3

2-

2 mol Cu2+ 2 mol S2O3

2-

Pour into Volumetric flask

V = 250mlM = ?

25ml transfer

1g KI excess/ starch

10 ml HNO3

titrated

Water added 250ml

2.5g brass

32

2

232

2

1082.2).(22

0282.01.0).(

22

)()(

CuMole

CuMole

OSMVCuMV

M x V = Mol Cu 2+

M x V = 2.82 x 10-3 M x 0.025 = 2.82 x 10-3 M = 2.82 x 10-

3 0025 M = 1.13 x 10-3

M

Convert mole Cu → Mass Cu

X RMM

X 10

2Cu2+ + 4I- → I2 + 2CuI

I2 + 2S2O32- → S4O6

2- + 2I-

0.456 g brass react with 25ml HNO3 producing Cu2+ ions. Sol was titrate with 0.1M S2O3

2- and end point, reached when 28.5ml added. Cal mol, mass, molarity of Cu 2+ and %

Cu in brass.

Redox Titration - % Cu in Brass

Na2S2O3

M = 0.1MV = 28.5ml

I2

M = ?

Mole ratio (1 : 1)2 mol Cu2+ : 1 mol I2 : 2 mol S2O3

2-

2 mol Cu2+ 2 mol S2O3

2-

transfer

1g KI excess starch

25ml HNO3

titrated

0.456g brass

32

2

232

2

1085.2).(22

0285.01.0).(

22

)()(

CuMole

CuMole

OSMVCuMV

M x V = Mol Cu 2+

M x V = 2.85 x 10-3 M x 0.025 = 2.85 x 10-3 M = 2.85 x 10-

3 0025 M = 1.14 x 10-3

M

Convert mole Cu → Mass Cu

2.85 x 10-3 x 63.5 = 0.18 g Cu

X RMM

% Cu = mass Cu x 100% mass brass = 0.18 x 100% 0.456 = 39.7 %

% Calcium carbonate in egg shell - Back Titration

250ml, 2M HNO3

Amt of HNO3 added

Amt of base (egg)

Amt of HNO3 left

Titrate NaOH M = 1.0 V = 17.0ml

Amt HNO3 react = Amt HNO3 – Amt HNO3 add left

HNO3 left

Transfer to flask

Left overnight in acid

added

25 g of egg shell (CaCO3) dissolved in 250 ml, 2 M HNO3. Sol titrated with NaOH.

17 ml, 1M NaOH needed to neutralize excess acid. Cal % CaCO3 by mass in egg shell.

NaOH + HNO3 → NaNO3 +

H2O M = 1.00M mol = ?V = 17 ml

Amt HNO3 add = M x V = 2.0 x 0.250 = 0.50 mol

Amt HNO3 react = Amt HNO3 add – Amt HNO3 left = 0.50 – 1.7 x 10-2 = 0.483 mol 2HNO3 + CaCO3 → (CaNO3)2 + H2O + 2CO2 Mole Mole 0.483 ?

Mole ratio (2 : 1) 2 mol HNO3 - 1 mol CaCO3 0.483 mol HNO3 - o.242 mol CaCO3

2107.1..

11

).(017.000.111

acidMole

acidMole

VMVM

aa

bb

25 g impureCaCO3 in egg shell

Convert mole CaCO3 → Mass /g

X RMM

0.242 x 100 = 24.2 g CaCO3

% CaCO3 = mass CaCO3 x 100% mass egg = 24.2 x 100% 25.0 = 96.8 %

% Calcium carbonate in egg shell - Back Titration

Amt of HCI added

Amt of base (egg)

Amt of HCI left

Titrate NaOH M = 0.10 V = 23.8 ml

Amt HCI react = Amt HCI – Amt HCI add left

HCI left

Transfer to flask

Left overnight in acid

added

NaOH + HCI → NaCI + H2O M = 0.1 M mol = ?V = 23.8 ml

Amt HCI add = M x V = 0.2 x 0.272 = 0.0544 mol

Amt HCI react = Amt HCI add – Amt HCI left = 0.0544 – 2.38 x 10-3 = 0.00306 mol 2HCI + CaCO3 → CaCI3 + H2O + CO2 Mole Mole 0.00306 ?

Mole ratio (2 : 1) 2 mol HCI - 1 mol CaCO3 0.00306 mol HCI - o.00153 mol

CaCO3

31038.2..

11

).(238.01.011

acidMole

acidMole

VMVM

aa

bb

Convert mole CaCO3 → Mass /g

X RMM

0.00153 x 100 = 0.153 g CaCO3

% CaCO3 = mass CaCO3 x 100% mass egg = 0.153 x 100% 0.188 = 81.4 %

0.188g of egg shell (CaCO3) dissolved in 27.2 ml, 0.2 M HCI. Sol was titrated with NaOH. 23.8ml, 0.10M NaOH need to

neutralize excess acid. Cal mol, mass and % of CaCO3 by mass in egg shell.

0.188g impureCaCO3 in egg shell

27.20ml, 0.2M HCI

Amt of HCI added

Amt of base

Amt of HCI left

Titrate NaOH M = 0.1108 V = 33.64 ml

Amt HCI react = Amt HCI – Amt HCI add left

HCI left

Transfer to flask

Left overnight in acid

added

NaOH + HCI → NaCI + H2O M = 0.1108 M mol = ?V = 33.64 ml

Amt HCI add = M x V = 0.250 x 0.05 = 0.0125 mol

Amt HCI react = Amt HCI add – Amt HCI left = 0.0125 – 3.727 x 10-3 = 0.008773 mol 2HCI + Ca(OH)3 → CaCI3 + H2O Mole Mole 0.008773 ?

Mole ratio (2 : 1) 2 mol HCI - 1 mol Ca(OH)2 0.008773 mol HCI - o.004386 mol

Ca(OH)2

310727.3..

11

).(03364.01108.0

11

acidMole

acidMole

VMVM

aa

bb

Convert mole Ca(OH)2 → Mass /g

X RMM

0.004386 x 74.1 = 0.325g Ca(OH)2

% Ca(OH)2 = mass Ca(OH)2 x 100% mass impure = 0.325 x 100% 0.5214 = 62.3 %

50 ml, 0.250M HCI

% Calcium hydroxide in antacid tablet - Back Titration

0.5214 g of impure Ca(OH)2 from antacid was dissolved in 50 ml, 0.250M HCI.

33.64ml, 0.1108 M NaOH need to neutralize excess acid. Cal % Ca(OH)2 in tablet.

0.5214g impureCa(OH)2

Amt of NaOH added

Amt of acid

Amt of NaOH left

Titrate HCI M = 0.5

V = 17.6 ml

Amt NaOH react = Amt NaOH – Amt NaOH add left

NaOH left

Transfer to flask

Left overnight in acid

added

HCI + NaOH → NaCI + H2O M = 0.5 M mol = ?V = 17.6 ml

Amt NaOH add = M x V = 2 x 0.02 = 0.04 mol

Amt NaOH react = Amt NaOH add – Amt NaOH left = 0.04 – 8.8 x 10-3 = 0.0312 mol 2NaOH + H2A → Na3 A+ 2H2O Mole Mole 0.0312 ?

Mole ratio (2 : 1) 2 mol NaOH - 1 mol acid 0.0312 mol NaOH - 0.0156 mol acid

3108.8.

11

).(0176.05.011

baseMole

acidMole

VMVM

bb

aa

Molar mass of insoluble acid in tablet -Back Titration

2.04 g insoluble acid dissolve in 20 ml, 2M NaOH. Excess NaOH require

17.6 ml, 0.5M HCI to neutralize it. Find molar mass acid

2.04 g impureacid H2A

20 ml, 2M NaOH

Molar Mass Acid0.0156 mol acid - 2.04 g 1 mol acid - 2.04 0.0156 = 131

Acknowledgements

Thanks to source of pictures and video used in this presentation

Thanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/

Prepared by Lawrence Kok

Check out more video tutorials from my site and hope you enjoy this tutorialhttp://lawrencekok.blogspot.com