Question

Four solutions must be prepared, which should contain the following pH values: 1, 4, 9 and 12. The following chemicals are available:

- sulphuric Acid (96% m/m)

- Ethanoic acid (qpp%)

- Ammonia solution (25% (m/m))

- Sodium hydroxide platelets

A) Name the appropriate chemicals that can be used to prepare the four solutions. Write a reaction to explain the choice of the chemicals for each of the four solutions.

B) Calculate the required mass of the chemicals that is needed to make 250ml of the solution (total of 4 solutions must be calculated).

If it involves a buffer solution, assume that:

total concentration = [acid] + [base] = 1M

ethanoic acid 100%

Answer 1

Part A.

A solution with pH 1 can be prepared from sulphuric acid (H₂SO₄).

Explanation: H₂SO₄ is a strong acid, which has a pH range of 0-3.

A solution with pH 4 can be prepared from ethanoic acid or acetic acid (CH₃COOH)

Explanation: CH₃COOH is a weak acid, which has a pH range of 3-7.

A solution with pH 9 can be prepared from aqueous ammonia solution or ammonium hydroxide (NH₄OH)

Explanation: NH₄OH is a weak base, which has a pH range of 7-10.

A solution with pH 12 can be prepared from sodium hydroxide (NaOH).

Explanation: NaOH is a strong base, which has a pH range of 10-14.

Part B.

Solution with pH 1.

i.e. 1 = -Log[H⁺]

i.e. [H⁺] = 10^-1 M

[H₂SO₄] = (10^-1/2) M = 5*10^-2 M

No. of moles of H₂SO₄ required = 5*10^-2 mol/L * 0.25 L = 0.0125 mol

i.e. Mass of 100% m/m H₂SO₄ = 0.0125 mol * 98 g/mol = 1.225 g

Therefore, the mass of 96% m/m H₂SO₄ = 100*1.225/96 = 1.276 g

Solution with pH 4.

4 = 1/2 (pKa - Log[CH₃COOH])

i.e. 8 = 4.74 - Log[CH₃COOH]

i.e. Log[CH₃COOH] = -3.26

i.e. [CH₃COOH] = 10^-3.26 M = 5.5*10^-4 M

No. of moles of CH₃COOH required = 5.5*10^-4 mol/L * 0.25 L = 1.374*10^-4 mol

i.e. The mass of 100% pure acetic acid = 1.374*10^-4 mol * 60 g/mol = 8.24*10^-3 g

Note: According to acetic acid (qpp%), you should follow the weight, check what is acetic acid (qpp%) and then proceed for the calculation accordingly.

Solution with pH 9.

9 = 14 - {1/2 (pKb - Log[NH₄OH])}

i.e. 4.74 - Log[NH₄OH] = 10

i.e. Log[NH₄OH] = -5.26

i.e. [NH₄OH] = 10^-5.26 M = 5.5*10^-6 M

No. of moles of NH₄OH required = 5.5*10^-6 mol/L * 0.25 L = 1.374*10^-6 mol

i.e. The mass of 100% m/m ammonia solution = 1.374*10^-6 mol * 35 g/mol = 4.81*10^-5 g

Therefore, the mass of 25% m/m ammonia solution = 4.81*10^-5 g * 100/25 = 1.92*10^-4 g

Solution with pH 12.

i.e. 12 = 14 - (-Log[OH^-])

i.e. [OH^-] = 10^-2 M

[NaOH] = 10^-2 M

No. of moles of NaOH required = 10^-2 mol/L * 0.25 L = 2.5*10^-3 mol

i.e. Mass of NaOH platelets = 2.5*10^-3 mol * 40 g/mol = 0.1 g