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3 parts
a

1.325 g\text{1.325 g}

d

24.62 cm3\text{24.62 cm}^3

e

0.102 mol dm−3\text{0.102 mol dm}^{-3}

Step-by-step solution

21 steps

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Part a· 3 steps

Calculate the mass of anhydrous sodium carbonate required to prepare 250 cm³ of 0.05 M solution.

  1. Calculate the molar mass of anhydrous sodium carbonate (Na₂CO₃).

    Given atomic masses: Na = 23, C = 12, O = 16.

    Molar mass of Na2CO3=(2×23)+12+(3×16)=46+12+48=106 g mol−1\text{Molar mass of Na}_2\text{CO}_3 = (2 \times 23) + 12 + (3 \times 16) = 46 + 12 + 48 = 106 \text{ g mol}^{-1}

  2. Calculate the moles of Na₂CO₃ required.

    Given volume (VV) = 250 cm³ = 0.250 dm³.
    Given concentration (CC) = 0.05 mol dm⁻³.

    Moles(n)=C×V=0.05 mol dm−3×0.250 dm3=0.0125 mol\text{Moles} (n) = C \times V = 0.05 \text{ mol dm}^{-3} \times 0.250 \text{ dm}^3 = 0.0125 \text{ mol}

  3. Calculate the mass of Na₂CO₃ required.

    Mass(m)=n×Molar mass=0.0125 mol×106 g mol−1=1.325 g\text{Mass} (m) = n \times \text{Molar mass} = 0.0125 \text{ mol} \times 106 \text{ g mol}^{-1} = 1.325 \text{ g}

    The mass of anhydrous sodium carbonate required is 1.325 g\boxed{\text{1.325 g}}.

Part b· 5 steps

Describe how to prepare the solution in a 250 cm³ volumetric flask.

  1. Weigh out 1.325 g of anhydrous sodium carbonate accurately using a weighing boat and an electronic balance.

  2. Transfer the weighed solid into a clean 100 cm³ beaker.

    Add about 50 cm³ of distilled water and stir with a glass rod until the solid completely dissolves.

  3. Carefully transfer the solution from the beaker into a 250 cm³ volumetric flask using a funnel.

    Rinse the beaker and the glass rod several times with small portions of distilled water, adding all washings to the volumetric flask.

  4. Add distilled water to the volumetric flask until the bottom of the meniscus just touches the 250 cm³ mark, ensuring your eye is at the same level as the mark.

  5. Stopper the volumetric flask and invert it several times to ensure the solution is thoroughly mixed and homogeneous.

Part c· 6 steps

Describe how to carry out the titration of 25.0 cm³ of the sodium carbonate solution with hydrochloric acid using methyl orange indicator.

  1. Rinse a burette with the hydrochloric acid solution and then fill it with the acid, ensuring the jet is filled and reading the initial volume at the bottom of the meniscus.

  2. Using a pipette, transfer 25.0 cm³ of the prepared sodium carbonate solution into a clean conical flask.

  3. Add 2-3 drops of methyl orange indicator to the sodium carbonate solution in the conical flask.

    The solution will turn yellow.

  4. Titrate the sodium carbonate solution by slowly adding hydrochloric acid from the burette, swirling the conical flask continuously.

    Add the acid drop by drop as you approach the endpoint.

  5. Stop adding acid when the indicator changes color from yellow to a permanent orange/peach color.

    Record the final burette reading.

  6. Repeat the titration several times (at least two more times) until concordant results (titre values within ±0.10 cm3\pm 0.10 \text{ cm}^3 of each other) are obtained.

Part d

24.62 cm3\text{24.62 cm}^3

Calculate the average volume of acid used.

Given titration results:
Titration 1: 24.80 cm³
Titration 2: 24.60 cm³
Titration 3: 24.65 cm³
Titration 4: 24.60 cm³

  1. Identify concordant results.

    Concordant results are those that are within ±0.10 cm3\pm 0.10 \text{ cm}^3 of each other.
    Titration 2 (24.60 cm³), Titration 3 (24.65 cm³), and Titration 4 (24.60 cm³) are concordant. Titration 1 (24.80 cm³) is not concordant with the others.

  2. Calculate the average volume using the concordant results.

    Average volume=24.60 cm3+24.65 cm3+24.60 cm33=73.85 cm33=24.6166... cm3\text{Average volume} = \frac{24.60 \text{ cm}^3 + 24.65 \text{ cm}^3 + 24.60 \text{ cm}^3}{3} = \frac{73.85 \text{ cm}^3}{3} = 24.6166... \text{ cm}^3

    Rounding to two decimal places, the average volume of acid used is 24.62 cm3\boxed{\text{24.62 cm}^3}.

Part e

0.102 mol dm−3\text{0.102 mol dm}^{-3}

Calculate the concentration of the hydrochloric acid.

  1. Write the balanced chemical equation for the reaction.

    Na2CO3(aq)+2HCl(aq)⟶2NaCl(aq)+H2O(l)+CO2(g)\text{Na}_2\text{CO}_3\text{(aq)} + 2\text{HCl(aq)} \longrightarrow 2\text{NaCl(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}

  2. Determine the mole ratio between Na₂CO₃ and HCl.

    From the balanced equation, 1 mole of Na₂CO₃ reacts with 2 moles of HCl.
    Mole ratio: Na2CO3:HCl=1:2\text{Na}_2\text{CO}_3 : \text{HCl} = 1 : 2.

  3. Calculate the moles of Na₂CO₃ used in the titration.

    Concentration of Na₂CO₃ = 0.05 mol dm⁻³
    Volume of Na₂CO₃ = 25.0 cm³ = 0.025 dm³

    Moles of Na2CO3=Concentration×Volume=0.05 mol dm−3×0.025 dm3=0.00125 mol\text{Moles of Na}_2\text{CO}_3 = \text{Concentration} \times \text{Volume} = 0.05 \text{ mol dm}^{-3} \times 0.025 \text{ dm}^3 = 0.00125 \text{ mol}

  4. Calculate the moles of HCl that reacted.

    Using the mole ratio:

    Moles of HCl=Moles of Na2CO3×2=0.00125 mol×2=0.00250 mol\text{Moles of HCl} = \text{Moles of Na}_2\text{CO}_3 \times 2 = 0.00125 \text{ mol} \times 2 = 0.00250 \text{ mol}

  5. Calculate the concentration of HCl.

    Average volume of HCl used = 24.62 cm³ = 0.02462 dm³

    Concentration of HCl=Moles of HClVolume of HCl=0.00250 mol0.02462 dm3=0.10154 mol dm−3\text{Concentration of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl}} = \frac{0.00250 \text{ mol}}{0.02462 \text{ dm}^3} = 0.10154 \text{ mol dm}^{-3}

    Rounding to three significant figures, the concentration of hydrochloric acid is 0.102 mol dm−3\boxed{\text{0.102 mol dm}^{-3}}.

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