Thursday 22 January 2015

7.3 Concentration of Acid and Alkali

Concentration
■ Concentration of a solution

The quantity [in gram or mol unit] of a dissolved solute in 1dm3 of the solution.
[Note: 1dm3=1000cm3]
Concentration of a solution [gdm-3]
= mass of solute (g)volume of solution( dm3)
Concentration of a solution [mol dm-3]
= number of mole of solute (mol)volume of solution(dm3)
Also known as Molarity [Unit: Molar or mol dm-3]

Example
Solution Meaning
5g dm-3 sodium hydroxide, NaOH 5g of sodium hydroxide, NaOH in 1dm3 of water
2 mol dm-3 lead(II) nitrate solution, Pb(NO3)2 2 mol of lead(II) nitrate solution, Pb(NO3)2 in 1dm3 of water
■ This video contains information on the concentration and molarity.


Worked-example 7.3(a)
50g of copper (II) sulphate anhydrous dissolved in water to produce a solution of 250cm3. Calculate the concentration of the solution produced in g dm-3?
Solution:
Step 1:
250cm3
=2501000dm3
= 0.25dm3

Step 2:
Concentration of copper (II) sulfate solution
=500.25
= 200g dm3
Worked-example 7.3(b)
Potassium chloride has a concentration of 14.9g dm-3. What is the concentration of this solution in mol dm-3? [Relative atomic mass: Cl, 35.5; K, 39]
Solution:
Step 1:
Molecular weight of potassium chloride, KI
= 39 + 35.5
= 74.5

Step 2:
Molarity of potassium chloride solution
=14.974.5
= 0.2mol dm3


Relationship between number of moles with molarity and volume of a solution.
■ Concentration of a solution

Number of moles =MV1000
Where M = molarity of solution(mol dm-3), V = volume of solution(cm3)
Reminder: Please take note on the unit before using this equation for calculation.

Worked-example 7.3(c)
100cm3hydrochloric acid contains 0.2 moles. Calculate the molarity of the hydrochloric acid.
Solution:
Number of moles =MV1000
0.2 =M×1001000
Molarity of hydrochloric acid, M =0.2×1000 100
= 2mol dm-3

Worked-example 7.3(d)
Calculate the number of moles of nitric acid in 200cm3 of 2 mol dm-3 of nitric acid solution.
Solution:
Number of moles =MV1000
Number of mol of nitric acid =2×2001000
= 0.4mol


Calculation involving concentration and molarity
Worked-example 7.3(e)
28g of potassium hydroxide dissolved in water to prepare 200cm3 solution. What is the molarity of potassium hydroxide produced? [Relative atomic mass: H, 1; O, 16, K, 39]
Solution:
Step 1:
Relative formula mass of KOH
= 39 + 16 + 1 = 56
Number of mole of KOH
= 2856
= 0.5mol
Step 2:
Number of moles =MV1000
0.5 =M×2001000
M =0.5×1000200
Molarity of potassium hydroxide, M = 2.5 mol dm-3

Worked-example 7.3(f)
Calculate the mass of calcium hydroxide contained in 50cm3 calcium hydroxide solution 0.1mol dm-3.
[Relative atomic mass: H, 1; O, 16; Ca, 40]

Solution:
Step 1:
Number of moles = MV1000
Number of moles of calcium hydroxide
= 0.1×50 1000
= 0.005mol
Step 2:
Relative formula mass of Ca(OH)2
= 40 + (16+1)×2 = 74
mass of calcium hydroxide = 0.005×74
= 0.37g


Preparation of a standard solution
■ A solution of known concentration

To prepare a standard solution it the desired molarity.
Volumetric flask with a known volume [example: 100cm3, 250cm3, 500cm3 and 1000cm3] must be used.
Mass in grams of solute required are weighed accurately
■ This video shows on the preparation of a standard solution.


Worked-example 7.3(f)
What is the mass of solid in grams required to prepare[Relative atomic mass: K, 39; I, 127]
(i) 250cm3 of a 1 M potassium iodide solution. (ii) 500cm3 of a 0.5 M potassium iodide solution.

Solution:
(i) 250cm3 of a 1 M potassium iodide solution.
Step 1:
Number of moles = MV1000
= 1.0×250 1000
= 0.25mol
Step 2:
Relative formula mass of KI
= 39 + 127 = 166
Mass of KI = 0.5×166
= 83g
(i) 500cm3 of a 0.5 M potassium iodide solution.
Step 1:
Number of moles = MV1000
= 0.5×500 1000
= 0.25mol
Step 2:
Relative formula mass of KI
= 39 + 127 = 166
Mass of KI = 0.25×166
= 83g

Laboratory Activity 7.3.1: Preparation of a standard solution
Preparation of a solution of certain concentration using the dilution method
■ Dilution

A process of diluting a concentrated solution by adding a solvent to obtain a diluted solution.

Number of moles of solute in the diluted solution = Number of moles of solution in the concentrated solution
Solution Before dilution After dilution
Volume V1 V2
Molarity M1 M2
Number of moles of solute V1M11000 V2M21000
Equation for dilution: V1M1 = V2M2
■ This video shows the procedure for performing a chemical dilution including the dilution equation.


Worked-example 7.3(h)
100cm3 of water is added to 150cm3 of 2mol dm-3 of KOH. Determine the molarity of the diluted solution.
Solution:
Step 1:
Total volume of solution
= 100 + 150 = 250cm3

Step 2:
M1V1 = M2V2
2(150) = M2(250)
M2 = 1.2mol dm-3

Worked-example 7.3(i)
What is the volume of 0.5mol dm-3 sulphuric acid that is required to be diluted with distilled water to produce 100cm3 of 0.1mol dm-3 solution of sulphuric acid?
Solution:
M1V1 = M2V2
0.5(V1) = 0.1(100)
V1 = 20cm3

Laboratory Activity 7.3.2: Preparation of solution of certain concentration using the dilution method


Relationship between the pH value and molarity of an acid and an alkali
■ pH value of acid and alkali depends on

Degree of dissociation

Molarity of the solution
■ For the same molarity solution

The higher the degree of dissociation in an acid, the lower the pH value.

The higher the degree of dissociation in an alkali, the higher the pH value.
■ For the same type of solution

The higher the molarity of an acidic solution, the lower is its pH value..

The higher the molarity of an alkaline solution, the higher is its pH value.

Laboratory Activity 7.3.3: Relationship between the pH value and molarity of an acid and an alkali


Solution to problem regarding molarity of acid and alkalis
■ Solution guidelines for problem regarding molarity of acid and alkalis

Write a balanced chemical equation.

List the information given and value needs to be determined.

Calculate the number of moles based on the information given.

Correlate the number of moles obtained with the number of moles of the substance in the equation.
■ Solution guidelines for problem regarding molarity of acid and alkalis

Number of moles = MV1000

Number of moles knowledge from Chapter 3

Worked-example 7.3(j)
25cm3 of nitric acid, HNO3, 2mol dm-3, reacted with excess zinc powder. Calculate the volume of hydrogen gas released under room conditions. [ Molar volume: 24dm3 at room conditions]
Solution:
Zn(s) + 2HNO3(aq) → Zn(NO3)2(aq) + H2(g)
Volume of acid, VA = 25cm3
Concentration of acid, MA = 2mol dm-3
Volume of hydrogen gas = ?
Number of moles of HNO3 = 2 ×251000=0.05 mol
From the equation, 2 mol of nitric acid releases 1 mol of hydrogen gas.
Thus 0.05 mol of nitric acid releases 0.025 mol of hydrogen gas.
Volume of hydrogen gas = 0.025×24 = 0.6dm3

Worked-example 7.3(k)
3g of magnesium, Mg, reacted completely with nitric acid, HNO3, 2mol dm-3 . Calculate the volume of acid used [Relative atomic molar: Mg, 24]
Solution:
Mg(s) + 2HNO3(aq) → Mg(NO3)2(aq) + H2(g)
Volume of acid, VA = ? cm3
Concentration of acid, MA = 2mol dm-3
Number of moles of magnesium = 3 24=0.125 mol
From the equation, 1 mol of magnesium react with 2 mol of nitric acid.
Thus, 0.125 mol of magnesium react with 0.25 mol of nitric acid.
MV 1000=number of moles
2 ×V1000=0.25
V=0.25x10000.2=125cm3

Worked-example 7.3(l)
Calculate the molarity of 50cm3 of hydrochloric acid, HCl, which reacted completely with 3.25 g zinc. [ Relative atomic molar: Zn, 65]
Solution:
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
Volume of acid, VA = 50cm3
Concentration of acid, MA = ? mol dm-3
Number of moles of Zn = 3.25 65=0.05 mol
From the equation, 1 mol of zinc react with 2 mol of hydrochloric acid.
Thus, 0.05 mol of zinc react with 0.1 mol of hydrochloric acid.
MV 1000=number of moles
M ×501000=0.1
M=0.1x100050=2mol dm3


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