Titration for IA (DCP) assessment

Acid Base Titration

Standardization HCI with primary std Na2CO3

Click here for expt 4.2

Standardization NaOH with primary std KHP

Click here or here for expt`

Titration bet NaOH with std HCI

Click here for expt 4.2a

Titration bet HCI with std NaOH

Click here for expt 4.2a

Determining water crystallization in hydrated Na2CO3 with std HCI

Click here for expt 4.4

Standardization KMnO4 with std ammonium

iron(II) sulphate Click here for expt 4.5

Iron (II) determination with std KMnO4

Click here for expt 4.6

Hypochlorite (OCI-) in bleach with iodine/thiosulphate

Click here for expt 4.8

Determining ethanoic acid in vinegar with std NaOH

Click here for expt 4.3

Copper(II) determination in brass

with iodine/thiosulphate Click here or here for expt` Click here for more expt

Standardization KI/I2 with std KIO3

Click here for expt 4.7 Click here for more expt

Determining acetylsalicylic acid in aspirin with std NaOH

Click here or here for expt` Click here for more expt

Vit C determination with iodine/thiosulphate Click here or here for expt

Click here more detail expt

Standardization Expt Acid/Base Titration Expt

Standardization Expt

Redox Titration Expt

Redox Titration

Standardization KI/I2 with std

sodium thiosulphate Click here for expt 4.7 Iodine/thiosulphate (iodometric titration)

Titration

Redox Titration Acid Base Titration Complexometric titration

Neutralization

Condition for Acid/Alkali Titration -One reactant – must be standard (known conc) or capable being standardised - Equivalent point – equal amt neutralize each other - End point measurable/detectable by colour change (indicator), pH change /conductivity

Acid/Base used as primary standard -Stable/solid

-Soluble in water -Does not decompose over time

Primary standard acids - Potassium hydrogen phthalate

Primary standard bases - Anhydrous sodium carbonate

10.6 gNa2CO3

Standard 0.1M Na2CO3

10.6g in 1 L

Volumetric flask Burette

Accurate known conc

Unable to prepare accurate conc of NaOH/HCI due to •Hygroscopic nature NaOH – absorb water vapour •HCI is in vapour state – difficult to measure amt

Volumetric flask Burette

Standard 0.1M KHP

20.4 g KHP 20.4 g in 1L

Unknown

Conc NaOH

Unknown

Conc HCI

? ?

Accurate known conc

Conc HCI=(0.211±0.001)M (Absolute uncertainty)

Standardization of (ACID) with standard (BASE)

Vol Na2CO3

Average vol Na2CO3 = 26.4 + 26.4 = 26.4cm3

2

Vol Na2CO3

Fin vol = (29.50 ± 0.05) Ini vol = (3.10 ± 0.05) Na2CO3 vol = (26.40 ± 0.10)

Uncertainty in vol Na2CO3 Add absolute uncertainty for final + initial = (0.05 + 0.05) = ± 0.10

Average Vol Na2CO3 ± uncertainty = 26.4 + 26.4 = (26.4± 0.10)cm3

2

Data Collection Data Processing

Error Analysis

% Uncertainty - burette

Absolute uncertainty vol x 100% Average vol added = 0.10 x 100% 26.4 = 0.38%

% Uncertainty - pipette

± 0.03

Total % Uncertainty = % uncertainty burette + % uncertainty pipette = 0.38% + 0.12% = 0.5%

Conc HCI = 0.211 ± 0.5% ( % uncertainty)

Absolute uncertainty vol x 100% Average vol added = 0.03 x 100% 25.00 = 0.12%

0.5 x 0.211 = 0.001 100 (% Absolute uncertainty)

Conc ± uncertainty % Error

Lit value - HCI = 0.200M Expt value – HCI = 0.211M Difference = 0.011 % Error – Difference x 100% Literature value 0.011 x 100% = 5.5%

0.200

Na2CO3

HCI

Standardization of (BASE) with standard (ACID)

Vol KHP

Average vol KHP = 26.4 + 26.4 = 26.4cm3

2

Vol KHP Fin vol = (29.50 ± 0.05) Ini vol = (3.10 ± 0.05) KHP vol = (26.40 ± 0.10)

Uncertainty in vol KHP Add absolute uncertainty for final + initial = (0.05 + 0.05) = ± 0.10

Average Vol KHP ± uncertainty = 26.4 + 26.4 = (26.4± 0.10)cm3

2

Data Collection Data Processing

Error Analysis

% Uncertainty - burette

Absolute uncertainty vol x 100% Average vol added = 0.10 x 100% 26.4 = 0.38%

% Uncertainty - pipette

± 0.03

Total % Uncertainty = % uncertainty burette + % uncertainty pipette = 0.38% + 0.12% = 0.5%

Conc NaOH = 0.106 ± 0.5% ( % uncertainty)

Absolute uncertainty vol x 100% Average vol added = 0.03 x 100% 25.00 = 0.12%

0.5 x 0.106 = 0.0005 = 0.001 100 (% Abs uncertainty)

Conc NaOH=0.106 ± 0.001M (Absolute uncertainty)

Conc ± uncertainty % Error

Lit value - NaOH = 0.100M Expt value– NaOH = 0.106M Difference = 0.006 % Error – Difference x 100% Literature value

0.006 x 100% = 6%

0.100

KHP

NaOH

NaOH

M = ? V = 25.0ml

KHP M = 0.100M V = 26.4 ml

KHP + NaOH → NaKP + H2O M = 0.100M M = ?

V = 26.40ml V = 25.0ml Mole ratio – 1: 1

Moles of Acid = MV = (0.100 x 0.0264) = 2.64 x 10-3 Mole ratio (1 : 1) • 1 mole acid neutralize 1 mole base • 2.64 x 10-3 acid neutralize 2.64 x 10-3 base Moles of Base = M x V = M x 0.025 M x 0.025 = 2.64 x 10-3

M = 0.106M

M aVa = 1 Mb Vb 1 0.1 x 26.40 = 1 M x 25.0 1

M b = 0.106M

HCI M = ? V = 25.0ml

Na2CO3 + 2HCI → 2NaCI + H2O + CO2

M = 0.100M M = ?

V = 26.4ml V = 25.0ml Mole ratio – 1: 2

Using mole ratio Using formula

Moles of Base = MV = (0.100 x 0.0264) = 2.64 x 10-3 Mole ratio (1 : 2) • 1 mole base neutralize 2 mole acid • 2.64 x 10-3 base neutralize 5.28 x 10-3 acid Moles of Acid = M x V = M x 0.025 M x 0.025 = 5.28 x 10-3

M = 0.211M

M bVb = 1 Ma Va 2 0.1 x 26.4 = 1 Ma x 25.0 2

Ma = 0.211M

Using formula Using mole ratio

Sample Titration Calculation

Calculation Calculation

Standardization of (BASE) with standard (ACID) Standardization of (ACID) with standard (BASE)

Na2CO3 M = 0.100M V = 26.4 ml

Click here Acid/Base calculation Click here Titration calculation

Video on Titration

Click here cal Na2CO3/HCI Click here cal NaOH/H2SO4

NaOH

M = ? V = 25.0ml

H2SO4

M = 1.00M V = 26.5cm3

2NaOH + H2SO4 → Na2SO4 + 2H2O M = ? M = 1.00M

V = 25.0ml V = 26.5ml Mole ratio – 2: 1

Moles of Acid = MV = (1.00 x 0.0265) = 2.65 x 10-2 Mole ratio (1 : 2) • 1 mole acid neutralize 2 mole base • 2.65 x 10-2 acid neutralize 5.30 x 10-2 base Moles of Base = M x V = M x 0.025 M x 0.025 = 5.30 x 10-2 M = 2.12M

Mb Vb = 2 Ma Va 1 M x 25.0 = 2 1.0 x 26.5 1

Mb = 2.12M

H2SO4 M = 0.5M V = 30.0ml

2NH4OH + H2SO4 → (NH4)2SO4 + 2H2O M = 1.5M M = 0.5M

V = ? ml V = 30.0ml Mole ratio – 2: 1

Using mole ratio Using formula

Moles of Acid = MV = (0.5 x 0.030) = 1.5o x 10-2 Mole ratio (2 : 1) • 1 mole acid neutralize 2 mole base • 1.50 x 10-2 acid neutralize 3.00 x 10-2 base Moles of Base = M x V = 1.5 x V 1.5 x V = 3.00 x 10-2

Vb = 0.02 dm3 = 20ml

M bVb = 2 Ma Va 1 1.5 x Vb = 2 0.5 x 30.0 1

Vb = 20ml

Using formula Using mole ratio

Acid/Base Titration Calculation

Calculation Calculation

NH3 / NH4OH M = 1.5M V = ? ml

25.0 cm3 of NaOH of unknown conc require 26.5cm3 of 1.0M sulphuric acid for complete neutralization. Find its molarity of NaOH.

Find vol of 1.5M aq NH3 required to completely neutralize 30.0cm3 of 0.5M sulphuric acid

2 1

Na2CO3

2.65g

HCI M = 2.0M V = ? ml

Na2CO3 + 2HCI → 2NaCI + CO2 + H2O M = 0.5M M = 2.0M

V = 50ml V = ? ml Mole ratio – 1: 2

Moles of Base = Mass/M = (2.65 ÷ 106) = 2.5 x 10-2 Mole ratio (1 : 2) • 1 mole base neutralize 2 mole acid •2.5 x 10-2 base neutralize 5.0 x 10-2 acid Moles of Acid = M x V = 2.0 x V 2.0 x V = 5 x 10-2

V = 0.25 dm3 = 25cm3

MbVb = 1 Ma Va 2 0.5 x 50.0 = 1 2.0 x V 2

Va = 25cm3

HCI M = ? V = 25.0ml

Na2CO3 + 2HCI → 2NaCI + H2O + CO2

M = 0.200M M = ?

V = 10.0ml V = 25.0ml Mole ratio – 1: 2

Using mole ratio Using formula

Moles of Base = MV = (0.200 x 0.010) = 2.00 x 10-3 Mole ratio (1 : 2) • 1 mole base neutralize 2 mole acid • 2.00 x 10-3 base neutralize 4.00 x 10-3 acid Moles of Acid = M x V = M x 0.025 M x 0.025 = 4.00 x 10-3

M = 0.160M

MbVb = 1 Ma Va 2 0.2 x 10.0 = 1 Ma x 25.0 2

Ma = 0.160M

Using formula Using mole ratio

Acid/Base Titration Calculation

Calculation Calculation

Na2CO3 M = 0.200M V = 10.0ml

Find vol of 2.0M HCI needed to neutralize 2.65g of sodium carbonate (Na2CO3) in 50ml water.

3 10.0cm3 of 0.200M Na2CO3 needed 25.0cm3 of HCI for neutralization. Find molarity of HCI.

4

V = 50ml

Moles = Mass/M = 2.65 106 = 0.025 mol

Acid

Mass = 1.325g

NaOH M = ? M V = 27.52 ml

C2H2O4 + 2NaOH → Na2C2O4 + 2H2O Mass - 1.325g M = ?

Mole – 0.0105 V = 27.52ml Mole ratio – 1: 2

Moles of Acid = Mass/M = (1.325 ÷ 126.08) = 1.05 x 10-2 Mole ratio (1 : 1) • 1 mole acid neutralize 2 mole base • 1.05 x 10-2 acid neutralize 2.10 x 10-2 base Mole of Base = M x V = M x 0.02752 M x 0.02752 = 2.10 x 10-2

M = 0.764M

M aVa = 1 Mb Vb 2 0.0105 = 1 M x 0.02752 2

Mb = 0.764M

HCI M = ? V = 33.7ml

Na2CO3 + 2HCI → 2NaCI + H2O + CO2

M = 1.37M M = ?

V = 20.0ml V = 33.7ml Mole ratio – 1: 2

Using mole ratio Using formula

Moles of Base = MV = (1.37 x 0.020) = 2.74 x 10-2 Mole ratio (1 : 2) • 1 mole base neutralize 2 mole acid • 2.74 x 10-2 base neutralize 5.48 x 10-2 acid Moles of Acid = M x V = M x 0.0337 M x 0.0337 = 5.48 x 10-2

M = 1.63M

M bVb = 1 Ma Va 2 1.37 x 20.0 = 1 Ma x 33.7 2

Ma = 1.63M

Using formula Using mole ratio

Acid/Base Titration Calculation

Calculation Calculation

Na2CO3 M = 1.37M V = 20.0ml

Vol NaOH need to neutralize 1.325 g ethanedioic acid (C2H2O4.2H2O) is 27.52cm3, calculate molarity of NaOH.

5 33.7cm3 of HCI neutralizes 20cm3 of Na2CO3 of 1.37M. Find molarity of acid.

6

Moles = Mass/M = 1.325 126.08 = 0.0105 mol

Acid/Base Titration Calculation – Ethanoic acid determination in vinegar

7

CH3COOH

M = ? V = 29.2ml

NaOH M = 0.1M V = 25.0ml

NaOH + CH3COOH → CH3COONa + H2O

M = 0.1M M = ?

V = 25.0ml V = 29.2ml

V = 250ml

M = 0.0856

Mole NaOH = MV = (0.1 x 0.025) = 2.5 x 10-3 Mole ratio (1 : 1) • 1 mole NaOH react 1 mole acid

• 2.5 x 10-3 mole NaOH react 2.5 x 10-3 acid Mole acid = M V = M x 0.0292 M x 0.0292 = 2.5 x 10-3 acid M = 2.5 x 10-3 0.0292 M = 0.0856M

Mole ratio – 1: 1

M1 x V1 = M2 x V2 M1 x 25 = 0.0856 x 250 M1 = 0.0856 x 250 25 M1 = 0.856M

Diluted 25x

V = 25 ml

M = ?

25.0ml of conc vinegar (ethanoic acid) was diluted to a total vol of 250 in a volumetric flask. Diluted vinegar was transfer to a burette. 25.0ml, 0.1M NaOH is pipette into a conical flask, with indicator added. End point reached when average 29.2 ml of diluted vinegar added. Find its molarity.

Using mole ratio Using formula

M aVa = 1 Mb Vb 1 0.1 x 0.025 = 1 M x 0.0292 1

M = 0.0856M

1

2

3

Moles bef dilution = Moles aft dilution M1 V1 = M2V2

M1 = Ini molarity M2= Final molarity V1 = Initial vol V2 = Final vol

Video on determination of ethanoic acid in vinegar

Acid/Base Titration Calculation - Empirical formula determination for Na2CO3. x H2O

8

HCI

M = 0.100 M V = 48.8ml

Na2CO3 M = ? M V = 25.0ml

2HCI + Na2CO3 → 2NaCI +CO2 + H2O

M = 0.1M M = ?

V = 48.8ml V = 25.0ml

V = 1L

M = ?

25 ml

transfer

Mole HCI = MV = (0.1 x 0.0488) = 4.88 x 10-3 Mole ratio (2 : 1) • 2 mole HCI react 1 mole base

• 4.88 x 10-3 mole HCI react 2.44 x 10-3 base Mole base = M V = M x 0.0250 M x 0.0250 = 2.44 x 10-3 base M = 2.44 x 10-3 0.0250 M = 0.0976M

Mole ratio – 2: 1

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

Diuted to 1L

27.82g

Na2CO3. xH2O

27.82g of hydrated (Na2CO3 . x H2O) dissolved in water, making up to 1L. 25.0ml of solution was neutralized by 48.8ml of 0.100M HCI. Cal molarity and mass of Na2CO3 present in 1L of solution. Find x .

mol/dm3 ↔ g/dm3

mol/dm3 x M → g/dm3 0.0976 x 106 = 10.36g/dm3

Convert moles to mass in 1L

Using mole ratio Using formula

M aVa = 2 Mb Vb 1 0.1 x 0.0488 = 2 M x 0.0250 1

M = 0.0976M

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 formula Na2CO3 . 10 H2O

1

2

3

4

5

Video on Na2CO3 calculation

Redox Titration Calculation- % Iron in iron tablet

Iron tablet contain hydrated iron (II) sulphate (FeSO4.7H2O). One tablet weighing 1.863g crushed, dissolved in water and solution made up to total vol of 250ml. 10ml of this solution added to 20ml of H2SO4 and titrated with 0.002M KMnO4. Average 24.5ml need to reach end point. Cal % iron(II) sulphate in iron tablet.

9

1.863 g

250ml

KMnO4 M = 0.002M V = 24.5 ml

Fe2+

M = ? V = 30ml

MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe2+ + 4H2O M = 0.002M M = ?

V = 24.5ml Mole ratio – 1: 5

Using mole ratio

Mole KMO4- = MV

= (0.002 x 0.0245) = 4.90 x 10-5 Mole ratio (1 : 5) • 1 mole KMO4

- react 5 mole Fe2+

• 4.90 x 10-5 KMO4-react 2.45 x 10-4 Fe2+

M aVa = 1 Mb Vb 5 0.002 x 0.0245 = 1 Moles Fe2+ 5

Moles = 2.45 x 10-4 Fe2+

Mass of (expt yield) = 1.703g Mass of (Actual tablet) = 1.863g % Fe in iron tablet = 1.703 x 100% 1.863 = 91.4%

Mole Mass Mole x RMM = Mass FeSO4

6.125 x 10-3 x 278.05 = 1.703g FeSO4

Using formula

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+ FeSO4.7H2O FeSO4 + 7H2O 1 mol 1 mol + 7 mol FeSO4 Fe2+ + SO4 2-

1 mol 1mol + 1mol 6.125 x 10-3 mol 6.125 x 10-3 mole Fe2+

1

2

3

4

Video on % Iron in iron tablet

Video on Fe2+/KMnO4 titration calculation

2 mol 1 mol 2CIO- + 2I- + 2H+ → I2 + 2CI- + H2O I2 + 2S2O3

2- → S4O62- + 2I-

1 mol 2 mol

10.0ml of bleach (CIO-) diluted to a total vol of 250ml. 20.0ml is added to 1g of KI (excess) and iodine produced is titrated with 0.0206M Na2S2O3.Using starch indicator, end point was 17.3ml. Cal molarity of CIO in bleach.

Redox Titration Calculation – CIO- in Bleach

10

Na2S2O3

M = 0.0206M V = 17.3ml

I2

M = ?

2S2O32- + I2 → S4O6

2- + 2I-

M = 0.0206 Mole = ?

V = 17.3ml V = 0.02

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

2-

2 mole CIO- 2 mole S2O32-

10.0ml CIO- transfer

V = 250ml

M = 1.78 x 10-2 M

20ml transfer

1g KI excess

added

Mole S2O32- = MV

= (0.0206 x 0.0173) = 3.56 x 10-4 Mole ratio (2 : 1) • 2 mole S2O3

2- react 1 mole I2

• 3.56 x 10-4 S2O32--react 1.78 x 10-4 I2

Mole ratio – 2: 1

2CIO- + 2I- + 2H+ → I2 + 2CI- + H2O 2CIO- I2 Mole = ? Mole = 1.78 x 10-4

Mole ratio – 2: 1

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

• 3.56 x 10-4 CIO- 1.78 x 10-4 I2

Moles of CIO- = M x V

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

Mole bef dilution = Mole aft dilution M1 V1 = M2V2

M1 = Ini molarity M2= Final molarity V1 = Initial vol V2 = Final vol M1 V1 = M2 V2 M1 x 10 = 1.78 x 10-2 x 250 M1 = 1.78 x 10-2 x 250 10 M1 = 0.445M

Diuted 25x

V = 10

M = ?

titrated

Water added

till 250ml

1

Using direct formula

M V (CIO+) = 2 = 1 M V (S203

2-) 2 1 Moles of CIO+ = 1

0.0206 x 0.0173 1

Moles of CIO- = 3.56 x 10-4

2

3

4

5

6

Video on CIO- in bleach

M aVa (KIO3) = 1 Mb Vb (C6H8O6) 3 0.002 x 0.0255 = 1 Mole C6H8O6 3

Mole C6H8O6 = 1.53 x 10-4

Mole C6H8O6 = M x V

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

1 mol 3 mol KIO3 + 5KI + 6H+ → 3I2 + 6K+ + 3H2O 3C6H8O6 + 3I2

→ 3C6H6O6 + 6I- + 6H+

3 mol 3 mol

Iodometric titration was performed on Vit C, (C6H8O6). 25ml Vit C is titrated with 0.002M KIO3 from burette, using excess KI and starch. Average vol KIO3 is 25.5ml. Cal molarity of Vit C.

Redox Titration Calculation – Vitamin C quantification

11

KIO3

M = 0.002M V = 25.5ml

Vit C

M = ? V = 25ml

KIO3 + 5KI + 6H+ → 3I2

+ 3H2O + 6K=

M = 0.002M Mole = ?

V = 25.5ml

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

1 mol KIO3 3 mol C6H8O6

V = 25ml

M = ?

25ml transfer

1g KI excess + starch

added

Mole KIO3 = MV = (0.002 x 0.0255) = 5.10 x 10-5 Mole ratio (1 : 3) • 1 mole KIO3 produce 3 mole I2

• 5.10 x 10-5 KIO3 produce 1.53 x 10-4 I2

Mole ratio – 1: 3

3C6H8O6 + 3I2 → 3C6H6O6 + 6I- + 6H+

Mole = ? 1.53 x 10-4

Mole ratio – 3: 3

Mole ratio (1 : 3) • 1 mol KIO3 react 3 mol C6H8O6

• 5.10 x 10-5 KIO3 react 1.53 x 10-4 C6H8O6

Mole C6H8O6 = M x V

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

titrated

Using mole ratio Using formula

Using formula

Vitamin C

1

2

3

4

2 mol 1 mol 2Cu2+ + 4I- → I2 + 2CuI I2 + 2S2O3

2- → S4O62- + 2I-

1 mol 2 mol

2.5g brass react with 10ml conc HNO3 producing Cu2+ ions. Solution made up to 250ml using water in a volumetric flask. Pipette 25ml of solution into conical flask. Na2CO3 added to neutralize excess acid. 1g KI (excess) and few drops of starch added to flask. Titrate with 0.1M S2O3

2- and end point, reached when 28.2ml added. Find molarity copper ions and % copper found in brass.

Redox Titration Calculation - % Cu in Brass

12

Na2S2O3

M = 0.1M V = 28.2ml

I2

M = ?

2S2O32- + I2 → S4O6

2- + 2I-

M = 0.1M Mole = ?

V = 28.2ml

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

2-

2 mole Cu2+ 2 mole S2O32-

Pour into

Volumetric flask

V = 250ml

M = ?

25ml transfer

1g KI excess + starch

added

Mole S2O32- = MV

= (0.1 x 0.0282) = 2.82 x 10-3 Mole ratio (2 : 1) • 2 mole S2O3

2- react 1 mole I2

• 2.82 x 10-3 S2O32-- react 1.41 x 10-3 I2

Mole ratio – 2: 1

2Cu2+ + 4I- → I2 + 2CuI Mole = ? 1.41 x 10-3 I2

Mole ratio – 2: 1

Mole ratio (2 : 1) • 2 mole Cu2+ 1 mole I2

• 2.82 x 10-3 Cu2+ 1.41 x 10-3 I2

Mole of Cu2+ = M x V

M x V = 2.82 x 10-3 M x 0.025 = 2.82 x 10-3 M = 2.82 x 10-3 0.025 M = 1.13 x 10-1 M Mass Cu = Molarity Cu x M Mass Cu = (1.13 x 63.5)g Cu in 1000ml = 7.18g Cu in 1000ml = 1.79g Cu in 250ml

10 ml HNO3

titrated

Water added

till 250ml

2.5g brass

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

Using formula Using mole ratio

Using formula

M V (Cu2+) = 2 = 1 M V (S203

2-) 2 1 Moles of Cu2+ = 1

0.1 x 0.0282 1

Moles of Cu2+ = 2.82 x 10-3

1

2

3

4 5

6

Titration

Redox Titration

Acid Base Titration

Complexometric Titration

Condition

Direct Titration Back Titration

• Both titrant and analyte soluble • Reaction is fast

H2SO4

HCI HNO3

NaOH NH4OH KOH Ba(OH)2

LiOH

Known conc /vol

of acid used

Amt of known

acid added

Amt of base (solid)

react with acid

Amt of excess

acid left

Titrated with known

conc/vol alkali

Amt base(solid) = Amt Known – Amt excess acid added acid left

Excess

acid left Transfer

to flask

known

Titrant - soluble

Analyte - soluble

Soluble

acid

Soluble base

(Alkali)

Condition

Left overnight in acid

CaCO3 in

egg shell

Impure limestone

unknown

added

• Sparingly soluble acid/base. • Rxn too slow, unable to dissolve/react • Calcium carbonate (egg shell) and impurities (base)

from antacid tablet/impure limestone. • Salicyclic acid in aspirin tablet.

Video on back titration calculation

Impure antacid

% Calcium carbonate in egg shell - Back Titration Calculation Back Titration

250.0ml, 2.0M

HNO3

Amt of

HNO3 added

Amt of base (egg)

Amt of

HNO3 left

Titrated with known

conc/vol NaOH

M = 1.0

V = 17.0ml

Amt HNO3 react = Amt HNO3 – Amt HNO3 with base add left

HNO3 left

Transfer

to flask

Left overnight in acid

25.0g impure

CaCO3 in egg shell

added

25.0g of egg shell (CaCO3) dissolved in 250.0ml, 2.0M HNO3. Solution was titrated with NaOH. 17.0ml, 1.00M NaOH needed to neutralize excess acid. Cal percentage CaCO3 by mass in egg shell.

NaOH + HNO3 → NaNO3 + H2O

M = 1.00M moles = ?

V = 17.0ml

Mole NaOH = MV = (1.00 x 0.017) = 1.7 x 10-2 Mole ratio (1 : 1) • 1 mole NaOH react 1 mole HNO3

• 1.7 x 10-2 mole NaOH react 1.7 x 10-2 HNO3

HNO3 left = 1.7 x 10-2 mol

Mole ratio – 1: 1

MbVb = 1 Ma Va 1 1.00 x 0.017 = 1 moles 1

mole HNO3 = 1.7 x 10-2

Using mole ratio Using formula

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

Amt HNO3 react = Amt HNO3 add – Amt HNO3 left with NaOH = 0.50 – 1.7 x 10-2 = 0.483 mol

2HNO3 + CaCO3 → (CaNO3)2 + H2O + 2CO2 Mole Mole

0.483 ? Mole ratio – 2: 1

Mole ratio (2 : 1) • 2 mol HNO3 react 1 mol CaCO3

• 0.483 mol HNO3 react o.242 mol CaCO3

Mass = Mole CaCO3 x RMM CaCO3 = 0.242 x 100 = 24.2g

% by mass = mass CaCO3 x 100% CaCO3 mass impure = (24.2/25.0) x 100% = 96.8%

1

2

3

4

5

6

7

8

3

Amt HNO3 add

Amt HNO3 left

Amt HNO3 react

with base

% Calcium hydroxide in an impure antacid tablet - Back Titration Calculation Back Titration

50.0ml, 0.250M

HCI

Amt of HCI added

Amt of base (solid)

Amt

HCI left

Titrated with known

conc/vol NaOH

M = 0.1108

V = 33.64ml

Amt HCI react = Amt HCI – Amt HCI with NaOH add left

HCI left

Transfer

to flask

Left overnight in acid

0.5214g impure

Ca(OH)2

added

0.5214g of impure Ca(OH)2 from antacid was dissolved in 50.0ml, 0.250M HCI. 33.64ml, 0.1108M NaOH needed to neutralize excess acid. Cal percentage sample containing Ca(OH)2 in a tablet.

NaOH + HCI → NaCI + H2O

M = 0.1108M moles = ?

V = 33.64ml

Mole NaOH = MV = (0.1108 x 0.03364) = 3.727 x 10-3 Mole ratio (1 : 1) • 1 mole NaOH react 1 mole HCI

• 3.727 x 10-3 mole NaOH react 3.727 x 10-3 HCI HCI left = 3.727 x 10-3 mol

Mole ratio – 1: 1

MbVb = 1 Ma Va 1 0.1108 x 0.03364 = 1 moles 1

mole HCI = 3.727 x 10-3

Using mole ratio Using formula

Amt HCI add = M V = 0.250 x 0.050 = 0.01250 mol

Amt HCI react = Amt HCI add – Amt HCI left with NaOH = 0.01250 – 3.727 x 10-3 = 0.008773 mol

2HCI + Ca(OH)2 → CaCI2 + 2H2O

Mole Mole

0.008773 ? Mole ratio – 2: 1

Mole ratio (2 : 1) • 2 mol HCI react 1 mol Ca(OH)2

• 0.008773 mol HCI react o.oo4386 mol Ca(OH)2

Mass = Mole Ca(OH)2 x RMM Ca(OH)2= 0.004386 x 74.1 = 0.3250g

% by mass = mass Ca(OH)2 x 100% Ca(OH)2 mass impure = (0.3250/0.5214) x 100% = 62.3%

1

2

3

4

5

6

7

8

3

Amt HCI Add

Amt HCI Left

Amt HCI react

with base

Molar mass of insoluble acid in a tablet -Back Titration Calculation Back Titration

20.0ml, 2.00M

NaOH

Amt NaOH added

Amt of acid (solid)

Amt of

NaOH left

Titrated with known

conc/vol HCI

M = 0.50

V = 17.6ml

Amt NaOH react = Amt NaOH – Amt NaOH with acid add left

NaOH left

Transfer

to flask

Left overnight

2.04g impure

dibasic acid H2A

added

2.04g insoluble dibasic acid dissolve in 20.0ml, 2.00M NaOH. Excess NaOH require 17.6ml, 0.5M HCI to neutralize it. Find molar mass acid

HCI + NaOH → NaCI + H2O

M = 0.50M moles = ?

V = 17.6ml

Mole HCI = MV = (0.50 x 0.0176) = 8.8 x 10-3 Mole ratio (1 : 1) • 1 mol HCI react 1 mol NaOH

• 8.8 x 10-3 mol HCI react 8.8 x 10-3 NaOH NaOH left = 8.8 x 10-3 mol

Mole ratio – 1: 1

MaVa = 1 Mb Vb 1 0.50 x 0.0176 = 1 moles 1

mole HCI = 8.8 x 10-3

Using mole ratio Using formula

Amt NaOH add = M V = 2.00 x 0.02 = 0.040 mol

Amt NaOH = Amt NaOH – Amt NaOH react with acid add left = 0.040 – 8.8 x 10-3 = 0.0312 mol

2NaOH + H2A → Na2 A + 2H2O

Mole Mole

0.00312 ? Mole ratio – 2: 1

Mole ratio (2 : 1) • 2 mol NaOH react 1 mol H2 A

• 0.0312 mol NaOH react o.o156 mol H2A

Molar Mass 0.0156 mol H2A - 2.04g 1 mol H2A - 2.04 0.0156 = 131

1

2

3

4

5

6 7

3

Amt NaOH add

Amt NaOH Left

Amt NaOH react

with acid

Click here on titration simulation Click here on titration simulation

Video on Titration

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

Simulation on Titration

Click here titration NaOH /HCI Click here titration video