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From the information given above, calculate the: (i). Concentration of A in m/dm (ii). Concentration of B in moldm

Chemistry

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a) (i) The piece of apparatus used for measuring the portion of Na2CO3\text{Na}_{2}\text{CO}_{3} solution is a pipette.

a) (ii) Three errors in the result presented are:

  • The calculated volume for the rough titration (24.00 cm3−0.00 cm3=24.00 cm3\text{24.00 cm}^{3} - \text{0.00 cm}^{3} = \text{24.00 cm}^{3}) does not match the recorded volume (24.10 cm3\text{24.10 cm}^{3}).
  • The second titration reading (22.5 cm3\text{22.5 cm}^{3}) lacks precision; burette readings should be recorded to two decimal places (e.g., 22.50 cm3\text{22.50 cm}^{3}).
  • The average titre value calculation includes values that are not concordant (e.g., 22.75 cm3\text{22.75 cm}^{3} and 22.40 cm3\text{22.40 cm}^{3} differ by more than 0.10 cm3\text{0.10 cm}^{3}). Only concordant values should be averaged.

b) (i) The colour of phenolphthalein in the Na2CO3\text{Na}_{2}\text{CO}_{3} solution (which is basic) is pink.

b) (ii) Yes, phenolphthalein is a suitable indicator for the first equivalence point of this titration.

  • Reason: Na2CO3\text{Na}_{2}\text{CO}_{3} is a weak base, and HCl\text{HCl} is a strong acid. Phenolphthalein changes colour in the pH\text{pH} range of 8.2-10.0\text{8.2-10.0}, which corresponds to the pH\text{pH} at the first equivalence point where Na2CO3\text{Na}_{2}\text{CO}_{3} is converted to NaHCO3\text{NaHCO}_{3}.

c) (i) The acid would turn blue litmus paper red.

c) (ii) After the burette had been washed and rinsed with water, the final cleaning step should be to rinse it with the acid (HCl\text{HCl}) to be used before filling it.

c) (iii) The type of reaction demonstrated by the experiment is an acid-base neutralization reaction.

c) (iv) The apparatus used to measure the volume of:

  • HCl\text{HCl} solution is a burette.
  • Na2CO3\text{Na}_{2}\text{CO}_{3} solution is a pipette.

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a) (i) Concentration of A in mol/dm3\text{mol/dm}^{3}

  1. Calculate the molar mass of NaHSO4\text{NaHSO}_{4}.

    Molar mass of NaHSO4=Na+H+S+(4×O)\text{Molar mass of NaHSO}_{4} = \text{Na} + \text{H} + \text{S} + (4 \times \text{O}) =23 g/mol+1 g/mol+32 g/mol+(4×16 g/mol)= \text{23 g/mol} + \text{1 g/mol} + \text{32 g/mol} + (4 \times \text{16 g/mol}) =23 g/mol+1 g/mol+32 g/mol+64 g/mol= \text{23 g/mol} + \text{1 g/mol} + \text{32 g/mol} + \text{64 g/mol} =120 g/mol= \text{120 g/mol}
  2. Convert the mass concentration to molar concentration.

    Concentration of A (CA)=Mass concentrationMolar mass\text{Concentration of A (C}_{A}) = \frac{\text{Mass concentration}}{\text{Molar mass}} CA=10.6 g/dm3120 g/mol\text{C}_{A} = \frac{\text{10.6 g/dm}^{3}}{\text{120 g/mol}} CA=0.08833 mol/dm3\text{C}_{A} = \text{0.08833 mol/dm}^{3}

    The concentration of A is 0.0883 mol/dm3\boxed{\text{0.0883 mol/dm}^{3}}.

    a) (ii) Concentration of B in mol/dm3\text{mol/dm}^{3}

  3. Write down the balanced chemical equation and identify the mole ratio.

    NaHSO4(aq)+NaOH(aq)⟶Na2SO4(aq)+H2O(l)\text{NaHSO}_{4}\text{(aq)} + \text{NaOH(aq)} \longrightarrow \text{Na}_{2}\text{SO}_{4}\text{(aq)} + \text{H}_{2}\text{O(l)}

    From the equation, the mole ratio of NaHSO4\text{NaHSO}_{4} to NaOH\text{NaOH} is 1:1\text{1:1}.
    So, nA=1\text{n}_{A} = 1 and nB=1\text{n}_{B} = 1.

  4. List the given values.

    • Concentration of A (CA\text{C}_{A}) = 0.08833 mol/dm3\text{0.08833 mol/dm}^{3} (from part a(i))
    • Volume of A (VA\text{V}_{A}) = 22.50 cm3\text{22.50 cm}^{3}
    • Volume of B (VB\text{V}_{B}) = 25.0 cm3\text{25.0 cm}^{3}
    • Concentration of B (CB\text{C}_{B}) = ?
  5. Use the titration formula to calculate CB\text{C}_{B}.

    CAVAnA=CBVBnB\frac{\text{C}_{A}\text{V}_{A}}{\text{n}_{A}} = \frac{\text{C}_{B}\text{V}_{B}}{\text{n}_{B}} CB=CAVAnBVBnA\text{C}_{B} = \frac{\text{C}_{A}\text{V}_{A}\text{n}_{B}}{\text{V}_{B}\text{n}_{A}} CB=(0.08833 mol/dm3)×(22.50 cm3)×1(25.0 cm3)×1\text{C}_{B} = \frac{(\text{0.08833 mol/dm}^{3}) \times (\text{22.50 cm}^{3}) \times 1}{(\text{25.0 cm}^{3}) \times 1} CB=1.98742525.0\text{C}_{B} = \frac{\text{1.987425}}{\text{25.0}} CB=0.079497 mol/dm3\text{C}_{B} = \text{0.079497 mol/dm}^{3}

    The concentration of B is 0.0795 mol/dm3\boxed{\text{0.0795 mol/dm}^{3}}.

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