Titration Calculations

Tags: Chemistry molarity acids and bases pH titration calculation



Acid-base Titration Calculations
Key Concepts
acid-base reactions involve a proton transfer

the acid donates a proton to the base

acid-base reactions are also known as neutralisation reactions

acid + base -----> salt + water
acid A + base B -----> conjugate acid of base B + conjugate base of acid A
(Lowry-Brönsted theory)


H+ + OH- ------> H2O is the most general neutralisation reaction

Equivalence point is the point at which the moles of H+ is equal to the moles of OH-
An indicator is used to show the equivalence point during a titration

A titration involves the progressive addition of one reactant from a burette
(usually the acid), to a known volume of the other reactant in a conical flask
(usually the base)
Calculations
Write the balanced chemical equation for the reaction

Extract all the relevant information from the question

Check that data for consistency, for example, concentrations are usually given in M or mol L-1 but volumes are often given in mL. You will need to convert the mL to L for consistency. The easiest way to do this is to multiply the volume in mL x 10-3

Calculate the moles of reactant (n) for which you have both the volume(V) and concentration(M) : n = M x V

From the balanced chemical equation find the mole ratio known reactant : unknown reactant

Use the mole ratio to calculate the moles of the unknown reactant

From the volume(V) of unknown reactant and its previously calculated moles(n), calculate its concentration(M): M = n ÷ V
Examples
30 mL of 0.10M NaOH neutralised 25.0mL of hydrochloric acid.
Determine the concentration of the acid

Write the balanced chemical equation for the reaction
NaOH(aq) + HCl(aq) -----> NaCl(aq) + H2O(l)

Extract the relevant information from the question:
NaOH V = 30mL , M = 0.10M HCl V = 25.0mL, M = ?

Check the data for consistency
NaOH V = 30 x 10-3L , M = 0.10M HCl V = 25.0 x 10-3L, M = ?

Calculate moles NaOH
n(NaOH) = M x V = 0.10 x 30 x 10-3 = 3 x 10-3 moles

From the balanced chemical equation find the mole ratio
NaOH:HCl
1:1

Find moles HCl
NaOH: HCl is 1:1
So n(NaOH) = n(HCl) = 3 x 10-3 moles at the equivalence point

Calculate concentration of HCl: M = n ÷ V
n = 3 x 10-3 mol, V = 25.0 x 10-3L
M(HCl) = 3 x 10-3 ÷ 25.0 x 10-3 = 0.12M or 0.12 mol L-1

50mL of 0.2mol L-1 NaOH neutralised 20mL of sulfuric acid.
Determine the concentration of the acid

Write the balanced chemical equation for the reaction
NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2H2O(l)

Extract the relevant information from the question:
NaOH V = 50mL, M = 0.2M H2SO4 V = 20mL, M = ?

Check the data for consistency
NaOH V = 50 x 10-3L, M = 0.2M H2SO4 V = 20 x 10-3L, M = ?

Calculate moles NaOH
n(NaOH) = M x V = 0.2 x 50 x 10-3 = 0.01 mol

From the balanced chemical equation find the mole ratio
NaOH:H2SO4
2:1

Find moles H2SO4
NaOH: H2SO4 is 2:1
So n(H2SO4) = ½ x n(NaOH) = ½ x 0.01 = 5 x 10-3 moles H2SO4 at the equivalence point

Calculate concentration of H2SO4: M = n ÷ V
n = 5 x 10-3 mol, V = 20 x 10-3L
M(H2SO4) = 5 x 10-3 ÷ 20 x 10-3 = 0.25M or 0.25 mol L-1

25.0mL of 0.05M Ba(OH)2 neutralised 40.0mL of nitric acid.
Determine the concentration of the acid.

Write the balanced chemical equation for the reaction
Ba(OH)2(aq) + 2HNO3(aq) -----> Ba(NO3)2(aq) + 2H2O(l)

Extract the relevant information from the question:
Ba(OH)2 V = 25.0mL, M = 0.05M HNO3 V = 40.0mL, M = ?

Check the data for consistency
Ba(OH)2 V = 25.0 x 10-3L, M = 0.05M HNO3 V = 40.0 x 10-3L, M = ?

Calculate moles Ba(OH)2
n = M x V = 0.05 x 25.0 x 10-3 = 1.25 x 10-3 mol

From the balanced chemical equation find the mole ratio
Ba(OH)2 : HNO3
1 : 2

Find moles HNO3
Ba(OH)2 : HNO3 is 1 : 2
So n(HNO3) = 2 x n(Ba(OH)2) = 2 x 1.25 x 10-3 = 2.5 x 10-3 mol

Calculate concentration of HNO3: M = n ÷ V
n = 2.5 x 10-3moles, V = 40.0 x 10-3L
M = 2.5 x 10-3 ÷ 40.0 x 10-3 = 0.0625M or 0.0625 mol L-1



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