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A volume of 0.25dm³ of 0.1mol/dm³ sodium hydroxide (NaOH) was put into a flask and few drops of phenolphthalein indicator were added. A volume of 0.20 dm³ of dilute hydrochloric acid (HCl) was used to completely neutralize sodium hydroxide.

Chemistry

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Question 1:

a)

i) Electrolysis is the process of using electrical energy to drive a non-spontaneous chemical reaction, typically to decompose a compound.

ii) The electrolyte used in this process is a solution containing copper ions, such as copper(II) sulfate solution (CuSO4(aq)\text{CuSO}_4\text{(aq)}).

iii) The missing apparatus is a DC power supply (or battery).

b)

i) The copper electrode decreases in mass as copper atoms lose electrons and dissolve into the electrolyte as Cu2+\text{Cu}^{2+} ions.

ii) The iron object increases in mass as copper ions (Cu2+\text{Cu}^{2+}) from the electrolyte gain electrons and deposit as copper metal onto its surface.

c) Two reasons for electroplating iron objects are:

  • To prevent corrosion (rusting) of the iron object.
  • To improve the appearance of the iron object by giving it a shiny copper coating.

d) The half equation for the copper electrode (anode) is: Cu(s)→Cu2+(aq)+2e− \text{Cu(s)} \to \text{Cu}^{2+}\text{(aq)} + 2\text{e}^-

Question 12:

a) The method of neutralization used is titration.

b) Phenolphthalein indicator is added to signal the endpoint of the titration. It changes color (from pink in basic solution to colorless in acidic solution) when the neutralization reaction is complete, indicating that the equivalence point has been reached.

c) The chemical equation for the reaction between sodium hydroxide and hydrochloric acid is: NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(l) \text{NaOH(aq)} + \text{HCl(aq)} \to \text{NaCl(aq)} + \text{H}_2\text{O(l)}

d) Hydrochloric acid should be handled with care because it is a corrosive acid that can cause severe chemical burns to skin and damage to eyes.

e) Calculate the concentration of dilute hydrochloric acid:

  1. Calculate the moles of sodium hydroxide (NaOH\text{NaOH}).

    Moles of NaOH=Concentration×Volume\text{Moles of NaOH} = \text{Concentration} \times \text{Volume} Moles of NaOH=0.1 mol/dm3×0.25 dm3=0.025 mol\text{Moles of NaOH} = 0.1 \text{ mol/dm}^3 \times 0.25 \text{ dm}^3 = 0.025 \text{ mol}
  2. Determine the moles of hydrochloric acid (HCl\text{HCl}) required for neutralization.

    From the balanced chemical equation NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(l)\text{NaOH(aq)} + \text{HCl(aq)} \to \text{NaCl(aq)} + \text{H}_2\text{O(l)}, the mole ratio of NaOH\text{NaOH} to HCl\text{HCl} is 1:11:1.

    Moles of HCl=Moles of NaOH=0.025 mol\text{Moles of HCl} = \text{Moles of NaOH} = 0.025 \text{ mol}
  3. Calculate the concentration of dilute hydrochloric acid.

    Concentration of HCl=Moles of HClVolume of HCl\text{Concentration of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl}} Concentration of HCl=0.025 mol0.20 dm3=0.125 mol/dm3\text{Concentration of HCl} = \frac{0.025 \text{ mol}}{0.20 \text{ dm}^3} = 0.125 \text{ mol/dm}^3

    The concentration of dilute hydrochloric acid is 0.125 mol/dm3\boxed{\text{0.125 mol/dm}^3}.

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