BPGHS SJChO Acid-Base Equilibrium I

Multiple Select

1. (Lv. 1) Identify the conjugate acid-base pairs for the following reaction: HNO₂(aq) + CN^-(aq) --> HCN(aq) + NO₂^-(aq)

• HNO₂(aq) and NO₂^-(aq)

• HNO₂(aq) and HCN(aq)

• CN^-(aq) and HNO₂(aq)

• HCN(aq) and CN^-(aq)

Answer/Feedback

Just by observing the reaction species, HNO₂ and NO₂^- differs from each other by 1 proton, and the same can be said for that between HCN and CN^-.

Multiple Choice

2. (Lv. 2) Identify the conjugate acid-base pairs for the following reaction: H₂SO₄(aq) + HNO₃(aq) --> HSO₄^-(aq) + H₂NO₃⁺(aq)

• HSO₄^-(aq) and SO₄^2-(aq)

• HNO₃(aq) and H₂NO₃⁺(aq)

• H₂NO₃⁺(aq) does not exist due to octet violation and thus the reaction is impossible.

• This reaction is impossible as 2 acids cannot react together.

Answer/Feedback

Via observation of the reaction, the conjugate acid-base pairs are H₂SO₄(aq)-HSO₄^-(aq) and HNO₃(aq)-H₂NO₃⁺(aq). Such a reaction does exist, and this is actually the 1^st step for the nitration of benzene (an organic reaction). Might you suggest a possible structure for H₂NO₃⁺, given that it decomposes to NO₂⁺ and H₂O?

Multiple Choice

3. (Lv. 1) Calculate the pH of the following at 25^oC: A solution of 0.82moldm^-3 HNO₃(aq).

• 0.663

• 1.228

• 0.086

• Impossible, as pH scale is only from 1-14 from my 'O' Levels.

Answer/Feedback

Since HNO₃ is a strong acid, it will fully dissociate to give H⁺(aq) and NO₃^-(aq) ions. Therefore, pH = -log₁₀(0.82) = 0.086186147.

Thinking about it, is it possible for pH to be = 0? To make things even more absurd, is it possible for pH to be negative? Think about this and challenge your 'O' level teachers' preaching!

Multiple Choice

4. (Lv. 1) Calculate the pH of the following at 25^oC: A solution of 0.035moldm^-3 H₂SO₄(aq).

• 1.154

• 1.455

• 1.005

• 1.541

Answer/Feedback

Since H₂SO₄ is a strong acid, it will fully dissociate to give H⁺(aq) and SO₄^2-(aq) ions. Take care that H₂SO₄ is diprotic, so H₂SO₄(aq) --> 2H⁺(aq) + SO₄^2-(aq), and therefore, pH = -log₁₀(2x.035) = 1.15490196.

Multiple Choice

5. (Lv. 2) Calculate the pH of the following at 25^oC: A solution of 0.40gdm^-3 NaOH(aq).

• 2.0

• 8.5

• 12.0

• 13.6

Answer/Feedback

Since NaOH is a strong base, it will fully dissociate in water to give Na⁺(aq) and OH^-(aq) ions. Therefore, since M_r of NaOH = 23.0 + 16.0 + 1.0 = 40.0, then concentration in moldm^-3 = 0.40/40 = 0.01moldm^-3. Thus, pH = 14 - pOH: pH = 14 - (-log₁₀0.01) = 12.

Multiple Choice

6. (Lv. 2) Calculate the pH of the following at 25^oC: A 15cm³ solution of 0.008moldm^-3 NaOH, with 118mg of K₂O(s) added to it.

• 11.90

• 13.46

• 12.96

• 13.24

Answer/Feedback

Moles of OH^-(aq) from NaOH = 0.008x0.015 = 0.00012mol

K₂O + H₂O(l) --> 2KOH(aq) --> 2K⁺(aq) + 2OH^-(aq)

Thus, Moles of OH^-(aq) from KOH = [118x10^-3/(2x39.1+16.0)] x2 = 0.002505307mol

Total [OH^-(aq)] = (0.00012+.002505307)/0.015 = 0.175020466moldm^-3

pH = 14 - pOH = 14 - (-log₁₀0.175020466) = 13.24308884.

Do you think this answer is accurate? If no, what assumption was used?

Multiple Choice

7. (Lv. 2) Calculate the pH of the following at 25^oC: A solution formed by mixing 17.5cm³ of 0.020moldm^-3 H₂SO₄ with 23.6cm³ of 0.025moldm^-3 HNO₃(aq).

• 1.13

• 1.47

• 1.50

• 1.64

Answer/Feedback

Total moles of H⁺(aq) = 0.02x0.0175x2 + 0.025x0.0236 = 0.00129mol

Therefore, [H⁺(aq)] = 0.00129/(0.0175+0.0236) = 0.031386861moldm^-3.

pH = -log₁₀0.031386861 = 1.503252112.

Multiple Choice

8. (Lv. 3) Calculate the pH of the following at 25^oC: A solution formed by mixing 16.8cm³ of 0.008moldm^-3 HCl(aq) with 22.5cm³ of 0.005moldm^-3 Ba(OH)₂(aq).

• 12.27

• 9.98

• 11.36

• 10.55

Answer/Feedback

Total moles of H⁺(aq) = 0.008x0.0168 = 0.0001344mol (Limiting)

Total moles of OH^-(aq) = 0.005x0.0225x2 = 0.000225mol (Excess)

Therefore, [OH^-(aq)] = (0.000225-0.0001344)/(0.0168+0.0225) = 0.002305343moldm^-3.

pH = 14 - (-log₁₀0.002305343) = 11.36273565.

Multiple Choice

9. (Lv. 3) Calculate the pH of the following at 25^oC: A solution formed by mixing 24.8cm³ of 0.075moldm^-3 HBr(aq) with 30.5cm³ of 0.045moldm^-3 NaOH.

• 1.98

• 2.05

• 2.28

• 2.62

Answer/Feedback

Total moles of H⁺(aq) = 0.075x0.0248 = 0.00186mol (Excess)

Total moles of OH^-(aq) = 0.045x0.0305 = 0.0013725mol (Limiting)

Therefore, [H⁺(aq)] = (0.00186-0.0013725)/(0.0248+0.0305) = 0.008815551moldm^-3.

pH = 14 - (-log₁₀0.008815551) = 2.054750511.

Multiple Choice

10. (Lv. 3) Calculate the pH of the following at 25^oC: A solution formed by mixing 18.6cm³ of 0.065dm^-3 H₂SO₄(aq) with 0.124g of Na₂O(s).

• 12.63

• 12.93

• 13.22

• 13.44

Answer/Feedback

Total moles of H⁺(aq) = 0.065x0.0186x2 = 0.002418mol (Limiting)

Na₂O(s) + H₂O(l) --> 2NaOH(aq)

Total moles of OH^-(aq) = [0.124/(23.1+23.1+16.0)] x2 = 0.003987138mol (Excess)

Therefore, [OH^-(aq)] = (0.003987138-0.002418)/(0.0186) = 0.084362272moldm^-3.

pH = 14 - (-log₁₀0.084362272) = 12.92614827

Multiple Choice

11. (Lv. 1) A 22.7cm³ solution of HNO₃(aq) has its pH determined by a pH meter at 25^oC. If the pH reading is 2.6, what is the moles of nitrate ions in the solution?

• 5.70x10^-5mol

• 0.111mol

• 0.553mol

• 2.27x10^-4mol

Answer/Feedback

[H⁺(aq)] = 10^-2.6 = [NO₃^-(aq)]

Therefore, moles of NO₃^-(aq) = 10^-2.6x0.0227 = 0.000057019mol

Multiple Choice

12. (Lv. 1) At 25^oC, a student weight out a sample of Barium Oxide, BaO(s). However, upon leaving the weighing scale, the student forgot what was the weight recorded. Assuming the BaO(s) is 100% pure, the student dissolved the BaO(s) in 25cm³ distilled water and measured the pH. The reading was 12.4. What was the mass of BaO(s) the student weighed out?

• 3.81x10^-13g

• 0.096g

• 0.048g

• 7.62x10^-13g

Answer/Feedback

pOH = 14 - 12.4 = 1.6

Thus, [OH^-(aq)] = 10^-1.6moldm-3

Moles of OH-(aq) = 10^-1.6 x 0.025 = 0.000627671mol.

BaO(s) + H₂O(l) --> Ba(OH)₂(aq) --> Ba²⁺(aq) + 2OH^-(aq)

Therefore, mass of BaO(s) = 0.000627671/2 x (137.3+16.0) = 0.048134023g

Multiple Choice

13. (Lv. 2) Calculate the pH for the following at 25^oC: A solution of 0.300moldm^-3 of propanoic acid, CH₃CH₂COOH (pK_a = 4.89).

• 2.71

• 4.89

• 3.77

• 5.12

Answer/Feedback

CH₃CH₂COOH(aq) --> CH₃CH₂COO^-(aq) + H⁺(aq)

From I.C.E. Table:

10^-4.89 = [H⁺(aq)][CH₃CH₂COO^-(aq)]/[CH₃CH₂COOH(aq)] = x²/0.3

x=0.001965896 (<5% of 0.3), so pH = -log₁₀0.001965896 = 2.706439373

Multiple Choice

14. (Lv. 2) At 25^oC, the pH of a solution of 0.35moldm^-3 Hydrofluoric Acid, HF(aq) is 1.82. Calculate the K_a of HF(aq) at this temperature.

• 6.85x10^-4moldm^-3

• 6.54x10^-4moldm^-3

• 6.22x10^-5moldm^-3

• 6.19x10^-5moldm^-3

Answer/Feedback

HF(aq) --> H⁺(aq) + F^-(aq)

[H+(aq)] = 10^-1.82 = [F^-(aq)]

From I.C.E. Table,

K_a = (10^-1.82)²/(0.35-0.015135612) = 0.000684118.

See here that the [H+(aq)] is >5% of 0.35, so we cannot take a shortcut and approximate. Please appreciate the fact that different weak acids are weak to different extents!

Multiple Choice

15. (Lv. 2) Calculate the pH for the following at 25^oC: A solution of 2.0moldm^-3 of ethylamine, CH₃CH₂NH₂ (K_b = 5.1x10^-4moldm^-3), which behaves like a weak base similar to NH₃.

• 9.81

• 10.43

• 11.01

• 12.50

Answer/Feedback

CH₃CH₂NH₂(aq) + H₂O(l) --> CH₃CH₂NH₃⁺(aq) + OH^-(aq)

From I.C.E. Table:

K_b = [CH₃CH₂NH₃⁺(aq)][OH^-(aq)]/[CH₃CH₂NH₂(aq)] = x²/2 = 5.1x10^-4moldm^-3.

[OH^-(aq)] = 0.031937438, pH = 14 - pOH = 14 - (-log₁₀0.031937438) = 12.50430009.

Multiple Choice

16. (Lv. 2) At 25^oC, the pH of a 0.200moldm^-3 sodium ethanoate solution is 9.02. Calculate the K_b of the ethanoate ion.

• 5.26x10^-10moldm^-3

• 5.48x10^-10moldm^-3

• 5.57x10^-10moldm^-3

• 5.82x10^-10moldm^-3

Answer/Feedback

CH₃COO^-(aq) + H₂O(l) --> CH₃COOH(aq) + OH^-(aq)

From I.C.E. Table,

K_b = (10^-(14-9.02))²/0.200 = 5.482390981x10^-10moldm^-3.