Chemistry Questions and Answers on Reactions

DEPARTMENT OF CHEMISTRY CHEM 1066 Introduction to Chemistry 1 Tutorial 2 Acid – Base Theory To be discussed: September 16-20, 2013. Solutions must be handed in at start of session.

1. For each of the following, identify the acids and bases (conjugates) in both the forward and reverse reactions. a) HClO2 + H2O Acid + base b) OCl + H2O Base + acid c) NH3 + H2PO4– Base + acid d) HCl + H2PO4 Acid + base e) HF + H2O Acid + base

– –

ClO2– + H3O+ conjugate base + conjugate acid HOCl + OH– conjugate acid + conjugate base NH4+ + HPO4 2– conjugate acid + conjugate base Cl– + H3PO4 conjugate base + conjugate acid F + H 3O + conjugate base + conjugate acid

2. In a laboratory experiment, students measured the pH of samples or rainwater and household ammonia. Determine: a) [H3O+] in the rainwater, with pH measured at 4.35 pH = -log10[H3O+] So log10 [H3O+] = -pH = -4.35 [H3O+] = 10-4.35 = 4.5 x 10-5 mol dm-3 b) [OH–] in the ammonia, with the pH measured at 11.28 pH + pOH = 14 11.28 + pOH = 14 pOH = 2.72 pOH = -log10[OH-] [OH-] = 1.9 x 10-3 mol dm-3

  Dr.

 R.

 Taylor/CHEM

 1066-­‐2013

3. The pH of a solution of HCl in water is found to be 2.50. What volume of water would you add to 1.00 L of this solution to raise the pH to 3.10? At pH = 2.5; pH = -log10[H3O+] = 2.5 So log10 [H3O+] = -pH = -2.5 [H3O+] = 10-2.5 = 3.2 x 10-3 mol dm-3 At pH = 3.1; pH = -log10[H3O+] = 3.1 So log10 [H3O+] = -pH = -3.1 [H3O+] = 10-3.1 = 7.9 x 10-4 mol dm-3 If solution is diluted from a [H3O+] = 3.2 x 10-3 mol dm-3 to [H3O+] = 7.9 x 10-4 mol dm-3 Then dilution factor = [H3O+] before/[H3O+] after = volume after/ volume before Therefore volume after = 4.05 L, hence volume added = 3.05 L 4. Calculate the [H3O+], [Cl–] and [OH–] in 0.015 M HCl(aq). Since HCl is a strong acid: complete ionization in water: HCl (aq) + H2O (l) → Cl– (aq) + H3O+ (aq) Moles of H3O+ (aq) = moles Cl– (aq) = moles HCl before reaction Therefore, [H3O+] = [Cl–] = 0.105 mol dm-3 Source of...