Chemistry Questions on Acids and Bases

(a) (i) Sodium hydrogen carbonate, NaHCO3, is a salt and will dissociate into ions when dissolved in water. Write an equation for this process.

Answer explanation

When sodium hydrogen carbonate (NaHCO3) dissolves in water, it dissociates into sodium ions (Na+) and bicarbonate ions (HCO3-). The equation is: NaHCO3 (s) → Na+ (aq) + HCO3- (aq).

(ii) One of the ions formed from the dissociation is amphiprotic because it can either accept or donate a proton. Write equations for each of these reactions. Acting as: an acid Acting as: a base

Answer explanation

The amphiprotic ion can act as an acid by donating a proton: HA ⇌ A- + H+. It can also act as a base by accepting a proton: A- + H+ ⇌ HA.

(c) The table below shows the concentration and pH of three basic solutions, sodium ethanoate, CH3COONa (aq), ammonia, NH3(aq), and sodium hydroxide, NaOH(aq). Explain why each of these solutions has a different pH value, yet they are the same concentration. Use equations to support your answer.

Answer explanation

Sodium ethanoate (CH3COONa) hydrolyzes to form CH3COO- and OH-, raising pH moderately. Ammonia (NH3) reacts with water to form NH4+ and OH-, increasing pH more. NaOH fully dissociates to Na+ and OH-, resulting in the highest pH.

(c) Compare the electrical conductivity of a hydrochloric acid solution, HCl(aq), with a solution of ethanoic acid, CH3COOH(aq), of the same concentration. Use equations to support your answer.

Answer explanation

HCl(aq) fully dissociates into H+ and Cl-, leading to high conductivity. CH3COOH(aq) partially dissociates (CH3COOH ⇌ H+ + CH3COO-), resulting in lower conductivity. Thus, HCl(aq) has higher electrical conductivity than CH3COOH(aq).

(a) Nitric acid, HNO3(aq), and ethanoic acid, CH3COOH(aq), are both acids. (i) Write equations to show their reactions with water, H2O(l). HNO3(aq) + H2O(l) CH3COOH(aq) + H2O(l) (ii) Use these equations to explain why they are classified as acids.

Answer explanation

HNO3(aq) + H2O(l) → H3O+(aq) + NO3−(aq) and CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO−(aq). Both produce H3O+ ions in solution, classifying them as acids.

(c) The table below provides information about solutions A to D. The solutions are known to be hydrochloric acid, HCl(aq), ammonium chloride, NH4Cl(aq), sodium hydroxide, NaOH(aq) and sodium chloride, NaCl(aq). (i) Identify solutions A to D. (ii) Justify your choices by comparing relative amounts of hydronium ion concentrations, [H3O+], in the solutions. Include relevant equations in your answer. (iii) Elaborate on the electrical conductivity of the four solutions.

Answer explanation

(i) A: HCl, B: NH4Cl, C: NaOH, D: NaCl. (ii) [H3O+] is highest in HCl (strong acid), lower in NH4Cl (weak acid), negligible in NaOH (strong base), and none in NaCl (neutral). (iii) HCl and NaOH are strong electrolytes, while NH4Cl is weak, and NaCl is neutral.

(b) (iv) Sodium ethanoate, CH3COONa is a salt. When solid sodium ethanoate is dissolved in water, it separates into ions. Use TWO relevant equations to explain whether the solution is acidic or basic.

Answer explanation

Sodium ethanoate dissociates into CH3COO- and Na+. The acetate ion (CH3COO-) can react with water: CH3COO- + H2O ⇌ CH3COOH + OH-. This produces OH- ions, making the solution basic.

(a) The hydrogensulfate ion, HSO4–, is an amphiprotic species because it can both accept or donate a proton, thus acting as an acid or base. Complete the equations for the reactions of the hydrogensulfate ion, HSO4–, with water in the box below. HSO4- acting as Equation an acid HSO4-(aq) + H2O(l) ⇌ a base HSO4-(aq) + H2O(l) ⇌

Answer explanation

HSO4– acts as an acid by donating a proton to water: HSO4–(aq) + H2O(l) ⇌ H3O+(aq) + SO4²–(aq). As a base, it accepts a proton: HSO4–(aq) + H2O(l) ⇌ HSO4²–(aq) + H3O+(aq).

(b) The pH and relative electrical conductivity of aqueous solutions of potassium hydroxide, KOH(aq), and ammonia, NH3(aq), are shown in the table below. Both have concentrations of 0.100 mol L–1. Chemical pH Conductivity KOH(aq) 13 good NH3(aq) 11.1 poor Explain the difference in pH and conductivity of these two solutions. Use relevant equations in your answer.

Answer explanation

KOH fully dissociates in water to produce OH- ions, resulting in a high pH (13) and good conductivity. NH3 partially ionizes to form NH4+ and OH-, leading to a lower pH (11.1) and poor conductivity due to fewer ions.

(a) Propanoic acid, C2H5COOH, is dissolved in water and the resulting solution has a pH of 4.2. (i) Complete the equation by writing the formulae of the two products. C2H5COOH(aq) + H2O(l) ⇌ ___________________ + ___________________ (ii) Explain the proton, H+, transfer in this reaction, and identify the two conjugate acid-base pairs.

Answer explanation

(i) C2H5COO- + H3O+. (ii) Propanoic acid donates a proton (H+) to water, forming hydronium (H3O+) and the propanoate ion (C2H5COO-). The conjugate acid-base pairs are (C2H5COOH, C2H5COO-) and (H2O, H3O+).