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Tutorial on Acid/Base, Redox, Back Titration

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NaOHM = ? V = 25.0ml

H2SO4

M = 1.00MV = 26.5cm3

2NaOH + H2SO4 → Na2SO4 + 2H2OM = ? M = 1.00MV = 25.0ml V = 26.5ml

Mole ratio – 2: 1

H2SO4

M = 0.5MV = 30.0ml

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

Mole ratio – 2: 1

Acid/Base Titration Calculation

Calculation Calculation

NH3 / NH4OHM = 1.5MV = ? 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

21

NaOHM = ? V = 25.0ml

H2SO4

M = 1.00MV = 26.5cm3

2NaOH + H2SO4 → Na2SO4 + 2H2OM = ? M = 1.00MV = 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 baseMoles of Base = M x V = M x 0.025 M x 0.025 = 5.30 x 10-2 M = 2.12M

Mb Vb = 2Ma Va 1M x 25.0 = 21.0 x 26.5 1Mb = 2.12M

H2SO4

M = 0.5MV = 30.0ml

2NH4OH + H2SO4 → (NH4)2SO4 + 2H2O M = 1.5M M = 0.5MV = ? 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 baseMoles 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 = 2Ma Va 11.5 x Vb = 20.5 x 30.0 1Vb = 20ml

Using formulaUsing mole ratio

Acid/Base Titration Calculation

Calculation Calculation

NH3 / NH4OHM = 1.5MV = ? 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

21

Na2CO3

2.65g

HCIM = 2.0MV = ? ml

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

Mole ratio – 1: 2

HCI M = ? V = 25.0ml

Na2CO3 + 2HCI → 2NaCI + H2O + CO2

M = 0.200M M = ?V = 10.0ml V = 25.0ml

Mole ratio – 1: 2

Acid/Base Titration Calculation

Calculation Calculation

Na2CO3

M = 0.200MV = 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

Na2CO3

2.65g

HCIM = 2.0MV = ? ml

Na2CO3 + 2HCI → 2NaCI + CO2 + H2OM = 0.5M M = 2.0MV = 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 acidMoles of Acid = M x V = 2.0 x V 2.0 x V = 5 x 10-2

V = 0.25 dm3 = 25cm3

MbVb = 1Ma Va 20.5 x 50.0 = 12.0 x V 2Va = 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 acidMoles of Acid = M x V = M x 0.025 M x 0.025 = 4.00 x 10-3

M = 0.160M

MbVb = 1Ma Va 20.2 x 10.0 = 1Ma x 25.0 2Ma = 0.160M

Using formulaUsing mole ratio

Acid/Base Titration Calculation

Calculation Calculation

Na2CO3

M = 0.200MV = 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

Titration

Example: Acid Base Titration

Condition

Direct Titration

• Both titrant and analyte soluble• Reaction is fast

H2SO4

HCIHNO3

NaOHNH4OHKOHBa(OH)2

LiOH

Titrant - soluble

Analyte - soluble

Soluble acid

Soluble base(Alkali)

And what if the base is insoluble in water ? Ex Ca(OH)2 (s)

Titration

Example: Acid Base Titration

Condition

Direct Titration Back Titration

• Both titrant and analyte soluble• Reaction is fast

H2SO4

HCIHNO3

NaOHNH4OHKOHBa(OH)2

LiOH

Known conc /vol of acid used

Titrant - soluble

Analyte - soluble

Soluble acid

Soluble base(Alkali)

Condition

CaCO3 inegg shell

Impure limestoneadded

• Sparingly soluble acid/base.• Reaction is SLOW• Ex: Calcium carbonate (egg shell) , calcium

carbonate in antiacid ( limestone) or calcium hydroxide from antacid table.

Impure antacid

Titration

example : Acid Base Titration

Condition

Direct Titration Back Titration

• Both titrant and analyte soluble• Reaction is FAST

H2SO4

HCIHNO3

NaOHNH4OHKOHBa(OH)2

LiOH

Known conc /vol

of acid used

Amt of excess acid left

Titrated with known conc/vol alkali

Excess acid left

Transfer to flask

Titrant - soluble

Analyte - soluble

Soluble acid

Soluble base(Alkali)

Condition

Left overnight in acid

CaCO3 inegg shell

Impure limestone

unknown

Amt of acid that reacts with insoluble base

Amt acid that has reacted = Amt Known – Amt excess acid added acid left

added

• Sparingly soluble acid/base.• Reaction is SLOW• Ex: Calcium carbonate (egg shell) , calcium

carbonate in antiacid ( limestone) or calcium hydroxide from antacid table.

known

Impure antacid

Ca(OH)2 + 2 HCl CaCl2 + 2H2O

% Calcium hydroxide in an impure antacid tablet - Back Titration CalculationBack 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 impureCa(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.

% Calcium hydroxide in an impure antacid tablet - Back Titration CalculationBack 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 Ca(OH )2 add left

HCI left

Transfer to flask

Left overnight in acid

0.5214g impureCa(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

Using mole ratio

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

Amt HCI react = Amt HCI add – Amt HCI left with Ca(OH)2 = 0.01250 – 3.727 x 10-3 = 0.008773 mol 2HCI + Ca(OH)2 → CaCI2

+ 2H2O

Mole Mole0.008773 ?

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 gfmCa(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%

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Amt HCI Add

Amt HCI Left

Amt HCI reactwith base