■ Concentration of a solution
► | The quantity [in gram or mol unit] of a dissolved solute in 1dm3 of the solution. [Note: 1dm3=1000cm3]
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► | Example
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✍ 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 = 0.25dm3 |
Step 2: Concentration of copper (II) sulfate solution = 200g dm3 |
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 = 0.2mol dm3 |
Relationship between number of moles with molarity and volume of a solution.
■ Concentration of a solution
► | 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:
= 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:
= 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 = 0.5mol |
Step 2: 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 of calcium hydroxide = 0.005mol |
Step 2: Relative formula mass of Ca(OH)2 = 74 = 0.37g |
Preparation of a standard solution
■ A solution of known concentration
► | To prepare a standard solution it the desired molarity.
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✍ 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: = 0.25mol |
Step 2: Relative formula mass of KI = 39 + 127 = 166 = 83g |
Step 1: = 0.25mol |
Step 2: Relative formula mass of KI = 39 + 127 = 166 = 83g |
Laboratory Activity 7.3.1: Preparation of a standard solution |
■ 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
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✍ 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 |
► | 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. |
► | 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. |
► | Number of moles = | |
► | 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 = 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 = 0.02524 = 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 = 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. |
✍ 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 = 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. |
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