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] $\text{=}\frac{\text{mass of solute (g)}}{\text{volume of solution}\left({\mathit{dm}}^{3}\right)}$ Concentration of a solution [mol dm-3] $\text{=}\frac{\text{number of mole of solute (mol)}}{\text{volume of solution}\left({\mathit{dm}}^{3}\right)}$ 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
 ► $\text{Number of moles =}\frac{MV}{1000}$ 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:
 $\text{Number of moles =}\frac{MV}{1000}$ $\text{0.2 =}\frac{M\mathrm{×}100}{1000}$ $\text{Molarity of hydrochloric acid, M =}\frac{\mathrm{0.2}\mathrm{×}1000}{100}$ = 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:
 $\text{Number of moles =}\frac{MV}{1000}$ $\text{Number of mol of nitric acid =}\frac{2\mathrm{×}200}{1000}$ = 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: $\text{Number of moles =}\frac{MV}{1000}$ $\text{0.5 =}\frac{M\mathrm{×}200}{1000}$ $\text{M =}\frac{\mathrm{0.5}\mathrm{×}1000}{200}$ 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: $\text{Number of moles =}\frac{MV}{1000}$ Number of moles of calcium hydroxide $\text{=}\frac{\mathrm{0.1}\mathrm{×}50}{1000}$ = 0.005mol Step 2: Relative formula mass of Ca(OH)2 $= 40 + \left(16+1\right)\mathrm{×}2$ = 74 $\text{mass of calcium hydroxide = 0.005}\mathrm{×}74$ = 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: $\text{Number of moles =}\frac{MV}{1000}$ $\text{=}\frac{\mathrm{1.0}\mathrm{×}250}{1000}$ = 0.25mol Step 2: Relative formula mass of KI = 39 + 127 = 166 $\text{Mass of KI = 0.5}\mathrm{×}166$ = 83g
(i) 500cm3 of a 0.5 M potassium iodide solution.
 Step 1: $\text{Number of moles =}\frac{MV}{1000}$ $\text{=}\frac{\mathrm{0.5}\mathrm{×}500}{1000}$ = 0.25mol Step 2: Relative formula mass of KI = 39 + 127 = 166 $\text{Mass of KI = 0.25}\mathrm{×}166$ = 83g

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

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.

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 = $\frac{2\mathrm{×}25}{1000}=0.05\mathit{mol}$ 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{×}$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. $\frac{MV}{1000}=\text{number of moles}$ $\frac{2\mathrm{×}V}{1000}=0.25$ $V\mathrm{=}\frac{0.25x1000}{0.2}\mathrm{=}125{\mathit{cm}}^{3}$

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. $\frac{MV}{1000}=\text{number of moles}$ $\frac{M\mathrm{×}50}{1000}=0.1$ $M\mathrm{=}\frac{0.1x1000}{50}\mathrm{=}2{\mathit{mol dm}}^{3}$