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Outline the procedure for the titration of a strong base (NaOH) with an acid (HX), explain why it was repeated, and calculate the concentration of both solutions.

Chemistry

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4 parts
c(i)

0.098 mol dm−30.098\, \text{mol dm}^{-3}

c(ii)

0.1106 mol dm−30.1106\, \text{mol dm}^{-3}

c(iii)

62.39 g/mol62.39\, \text{g/mol}

c(iv)
61.38

Step-by-step solution

9 steps

Part a

For the titration of a strong base (NaOH) with an acid (HX), a suitable indicator is **phenolphthalein

It changes color from colorless in acid to pink in alkaline solution at the endpoint.

Part b

The titration was repeated three times to obtain concordant results and ensure accuracy. Repeating the titration helps to minimize random errors and obtain a reliable average titre.

Part c(i)

0.098 mol dm−30.098\, \text{mol dm}^{-3}

Calculate the concentration of B (NaOH) in mol dm⁻³

  1. Calculate the molar mass of NaOH.

    Molar mass of NaOH=22.99 g/mol (Na)+15.999 g/mol (O)+1.008 g/mol (H)=39.997 g/mol≈40.0 g/mol\text{Molar mass of NaOH} = 22.99\, \text{g/mol (Na)} + 15.999\, \text{g/mol (O)} + 1.008\, \text{g/mol (H)} = 39.997\, \text{g/mol} \approx 40.0\, \text{g/mol}

  2. Calculate the moles of NaOH in 250 cm³ of solution.

    Moles of NaOH=Mass of NaOHMolar mass of NaOH=0.98 g40.0 g/mol=0.0245 mol\text{Moles of NaOH} = \frac{\text{Mass of NaOH}}{\text{Molar mass of NaOH}} = \frac{0.98\, \text{g}}{40.0\, \text{g/mol}} = 0.0245\, \text{mol}

  3. Calculate the concentration of NaOH in mol dm⁻³.

    Concentration of B (NaOH)=Moles of NaOHVolume of solution (dm3)=0.0245 mol0.250 dm3=0.098 mol dm−3\text{Concentration of B (NaOH)} = \frac{\text{Moles of NaOH}}{\text{Volume of solution (dm}^3\text{)}} = \frac{0.0245\, \text{mol}}{0.250\, \text{dm}^3} = \boxed{0.098\, \text{mol dm}^{-3}}

Part c(ii)

0.1106 mol dm−30.1106\, \text{mol dm}^{-3}

Calculate the concentration of A (HX) in mol dm⁻³

  1. Identify the concordant titres and calculate the average titre.

    The concordant titres are 22.20 cm³ and 22.10 cm³. The average titre given is 22.15 cm³.

    Average volume of A used=22.15 cm3=0.02215 dm3\text{Average volume of A used} = 22.15\, \text{cm}^3 = 0.02215\, \text{dm}^3

  2. Calculate the moles of NaOH used in the titration.

    Moles of NaOH=Concentration of B×Volume of B\text{Moles of NaOH} = \text{Concentration of B} \times \text{Volume of B} Moles of NaOH=0.098 mol dm−3×0.0250 dm3=0.00245 mol\text{Moles of NaOH} = 0.098\, \text{mol dm}^{-3} \times 0.0250\, \text{dm}^3 = 0.00245\, \text{mol}

  3. Determine the moles of HX from the stoichiometry.

    From the equation HX(aq)+NaOH(aq)⟶NaX(aq)+H2O(l)\text{HX}_{(\text{aq})} + \text{NaOH}_{(\text{aq})} \longrightarrow \text{NaX}_{(\text{aq})} + \text{H}_2\text{O}_{(\text{l})}, the mole ratio of HX : NaOH is 1:1.

    Moles of HX=Moles of NaOH=0.00245 mol\text{Moles of HX} = \text{Moles of NaOH} = 0.00245\, \text{mol}

  4. Calculate the concentration of HX.

    Concentration of A (HX)=Moles of HXVolume of A used (dm3)=0.00245 mol0.02215 dm3=0.1106 mol dm−3\text{Concentration of A (HX)} = \frac{\text{Moles of HX}}{\text{Volume of A used (dm}^3\text{)}} = \frac{0.00245\, \text{mol}}{0.02215\, \text{dm}^3} = \boxed{0.1106\, \text{mol dm}^{-3}}

Part c(iii)

62.39 g/mol62.39\, \text{g/mol}

Calculate the molar mass of A (HX)

  1. Use the given mass of HX in 1.0 dm³ and its calculated concentration.

    Molar mass of HX=Mass of HX in 1.0 dm3Concentration of A (HX)\text{Molar mass of HX} = \frac{\text{Mass of HX in 1.0 dm}^3}{\text{Concentration of A (HX)}} Molar mass of HX=6.90 g0.1106 mol dm−3=62.39 g/mol\text{Molar mass of HX} = \frac{6.90\, \text{g}}{0.1106\, \text{mol dm}^{-3}} = \boxed{62.39\, \text{g/mol}}

Part c(iv)61.38

Calculate the relative atomic mass of X

  1. Subtract the molar mass of hydrogen from the molar mass of HX.

    Relative atomic mass of X=Molar mass of HX−Molar mass of H\text{Relative atomic mass of X} = \text{Molar mass of HX} - \text{Molar mass of H} Relative atomic mass of X=62.39 g/mol−1.008 g/mol=61.38\text{Relative atomic mass of X} = 62.39\, \text{g/mol} - 1.008\, \text{g/mol} = \boxed{61.38}

Part d(i)

THREE precautions taken during the titration

  • Ensure the burette is filled below the eye level to avoid splashing and ensure accurate reading.
  • Read the burette at eye level to avoid parallax error.
  • Rinse the burette with the acid (solution A) and the pipette with the base (solution B) before use to prevent dilution or contamination.
  • Ensure no air bubbles are present in the burette tip.
  • Add the titrant dropwise near the endpoint and swirl the flask constantly.

Part d(ii)

TWO laboratory apparatus used for the titration

  • Burette (for dispensing the titrant, solution A)
  • Pipette (for accurately measuring a fixed volume of solution B)
  • Conical flask (for holding solution B and the indicator)
  • Retort stand and clamp (to hold the burette)
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