Question

4.How many mL of 0.25M NaOH are required to neutralize 50mL of 0.4M H2SO4?

6.Define: oxidation, reduction, anode, cathode, electrochemical half-cell, and the Standard Electrode Potential.

7.​You mix 500mL of 10-3M HCl with 500mL of 4 x 10-3M NaOH. What will be the pH of this solution?

8.​If 50mL of 10-2M HCl is diluted to a final volume of 800mL with distilled water, what will be the pH?

Answer 1

4. 2 moles of NaOH are needed to neutralise 1 mole of H₂SO₄.

number of moles of H₂SO₄ = Molarity * Volume = 50 * 10^-3 * 0.4 = 0.02 moles

Moles of NaOH = 0.02 * 2 = 0.04

Assuming a Volume V of NaOH: Moles = Molarity * Volume

0.25 * V = 0.04

V = 0.16 L = 160 ml

6. Oxidation: Oxidation is the process by which a species loses electrons.

    Reduction: Reduction is the process by which a species gains electrons.

    Anode: The electrode where Oxidation occurs.

    Cathode: The electrode where reduction occurs.

   electrochemical half-cell: Individual cathodes or anodes and their reactions constitute the electrochemical half- cell.

   Standard Electrode Potential: Standard Electrode Potential is defined by measuring the potential relative to a standard hydrogen electrode using a 1 Molar solution at 25 °C.

7. Number of moles of HCl = Molarity * Volume = 0.500 * 10^-3 = 5 * 10^-4 moles

   Number of moles of NaOH = 0.500 * 4 * 10^-3 * = 2 * 10^-3 moles

   Reaction equation:

   HCl + NaOH → NaCl + H₂O

We observe that 1 mole of HCl reacts with 1 mole of NaOH. This means that the entire 5 * 10^-4 moles of acid reacts with NaOH.

Remaining moles of NaOH = 2 * 10^-3 - 5 * 10^-4 = 1.5 * 10^-3 moles

We have 1.5 * 10^-3 moles of NaOH in a volume of 1 Litre.

[OH] = 1.5 * 10^-3 M

pOH = -log([OH^-]) = 2.83

pH = 14 - pOH = 14 - 2.83 = 11.17

8. On diluting, using the relation: M * V = constant

We have: 10^-2 * 50 = M * 800

Solving, M = 6.25 * 10^-4 M

[H⁺] = [HCl] = 6.25 * 10^-4 M

pH = -log([H⁺]) = 3.20