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] $\text{=}\frac{\text{mass of solute (g)}}{\text{volume of solution}({\mathit{dm}}^3)}$ Concentration of a solution [mol dm-3] $\text{=}\frac{\text{number of mole of solute (mol)}}{\text{volume of solution}({\mathit{dm}}^3)}$ 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 $\text{=}\frac{250}{1000}{\text{dm}}^3$ = 0.25dm3

Step 2: Concentration of copper (II) sulfate solution $\text{=}\frac{50}{0.25}$ = 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 $\text{=}\frac{14.9}{74.5}$ = 0.2mol dm3

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

►

$\text{Number of moles =}\frac{\mathrm{MV}}{1000}$ 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:

$\text{Number of moles =}\frac{\mathrm{MV}}{1000}$ $\text{0.2 =}\frac{M\mathrm{\times }100}{1000}$ $\text{Molarity of hydrochloric acid, M =}\frac{\mathrm{0.2}\mathrm{\times }1000}{100}$ = 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:

$\text{Number of moles =}\frac{\mathrm{MV}}{1000}$ $\text{Number of mol of nitric acid =}\frac{2\mathrm{\times }200}{1000}$ = 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 $\text{=}\frac{28}{56}$ = 0.5mol

Step 2: $\text{Number of moles =}\frac{\mathrm{MV}}{1000}$ $\text{0.5 =}\frac{M\mathrm{\times }200}{1000}$ $\text{M =}\frac{\mathrm{0.5}\mathrm{\times }1000}{200}$ 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: $\text{Number of moles =}\frac{\mathrm{MV}}{1000}$ Number of moles of calcium hydroxide $\text{=}\frac{\mathrm{0.1}\mathrm{\times }50}{1000}$ = 0.005mol

Step 2: Relative formula mass of Ca(OH)2 $\mathrm{=\; 40\; +\; (16+1)}\mathrm{\times }2$ = 74 $\text{mass of calcium hydroxide = 0.005}\mathrm{\times }74$ = 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: $\text{Number of moles =}\frac{\mathrm{MV}}{1000}$ $\text{=}\frac{\mathrm{1.0}\mathrm{\times }250}{1000}$ = 0.25mol

Step 2: Relative formula mass of KI = 39 + 127 = 166 $\text{Mass of KI = 0.5}\mathrm{\times }166$ = 83g

(i) 500cm³ of a 0.5 M potassium iodide solution.

Step 1: $\text{Number of moles =}\frac{\mathrm{MV}}{1000}$ $\text{=}\frac{\mathrm{0.5}\mathrm{\times }500}{1000}$ = 0.25mol

Step 2: Relative formula mass of KI = 39 + 127 = 166 $\text{Mass of KI = 0.25}\mathrm{\times }166$ = 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 $\frac{V_1{M}_1}{1000}$ $\frac{V_2{M}_2}{1000}$ 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 = $\frac{\mathrm{MV}}{1000}$
	►	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 = $\frac{2\mathrm{\times }25}{1000}=0.05\mathit{mol}$ 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$\mathrm{\times }$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 = $\frac{3}{24}=0.125\mathit{mol}$ 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. $\frac{\mathrm{MV}}{1000}=\text{number of moles}$ $\frac{2\mathrm{\times }\mathrm{V}}{1000}=0.25$ $V\mathrm{=}\frac{0.25x1000}{0.2}\mathrm{=}125{\mathit{cm}}^3$

✍ 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 = $\frac{3.25}{65}=0.05\mathit{mol}$ 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. $\frac{\mathrm{MV}}{1000}=\text{number of moles}$ $\frac{\mathrm{M}\mathrm{\times }50}{1000}=0.1$ $M\mathrm{=}\frac{0.1x1000}{50}\mathrm{=}2{\mathit{mol\; dm}}^3$

← 7.2 Strength of Acids and Alkalis

7.4 Neutralization →

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