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The image shows a titration experiment. Please provide the steps and calculations for: a) Average Volume of solution A used. b) Write a balanced equation & calculate the Conc of NaOH in mol dm⁻³. Ass

Chemistry

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To fully answer these questions, specific experimental data (titration readings, exact concentration of the standard acid, and which acid was used) is needed. Since this data is not provided in the image, I will outline the steps and use hypothetical values for the calculations to demonstrate the process.

  • a) Average Volume of solution A used.
  1. Record all accurate titre values (volumes of solution A used in each titration).

  2. Identify concordant titres.

    These are titre values that are within ±0.10 cm3\pm 0.10 \, \text{cm}^3 of each other.

  3. Calculate the average of the concordant titres.

    For example, if the concordant titres were 20.10 cm320.10 \, \text{cm}^3, 20.00 cm320.00 \, \text{cm}^3, and 20.05 cm320.05 \, \text{cm}^3:

    Average Volume=20.10+20.00+20.053 cm3\text{Average Volume} = \frac{20.10 + 20.00 + 20.05}{3} \, \text{cm}^3 Average Volume=60.153 cm3\text{Average Volume} = \frac{60.15}{3} \, \text{cm}^3 Average Volume=20.05 cm3\text{Average Volume} = \boxed{20.05 \, \text{cm}^3}

    (Note: This is a hypothetical example. You would use your actual experimental data.)

    • b) Write a balanced equation & calculate the Conc of NaOH in mol dm−3^{-3}.

    Assuming solution A (Aqueous sodium hydroxide, NaOH) is titrated against solution E (Hydrochloric Acid, HCl).

  4. Write the balanced chemical equation for the reaction.

    NaOH(aq)+HCl(aq)⟶NaCl(aq)+H2O(l)\text{NaOH(aq)} + \text{HCl(aq)} \longrightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}

    The mole ratio of NaOH to HCl is 1:11:1.

  5. State the known values and make hypothetical assumptions for unknown values.

    Let's assume:

    • Concentration of standard HCl (Solution E), CHCl=0.100 mol dm−3C_{\text{HCl}} = 0.100 \, \text{mol dm}^{-3}
    • Volume of HCl (Solution E) pipetted, VHCl=25.00 cm3=0.02500 dm3V_{\text{HCl}} = 25.00 \, \text{cm}^3 = 0.02500 \, \text{dm}^3
    • Hypothetical average volume of NaOH (Solution A) used, VNaOH=20.05 cm3=0.02005 dm3V_{\text{NaOH}} = 20.05 \, \text{cm}^3 = 0.02005 \, \text{dm}^3
    • Mole ratio from equation, nNaOH:nHCl=1:1n_{\text{NaOH}} : n_{\text{HCl}} = 1:1
  6. Calculate the moles of the known reactant (HCl).

    Moles of HCl=CHCl×VHCl\text{Moles of HCl} = C_{\text{HCl}} \times V_{\text{HCl}} Moles of HCl=0.100 mol dm−3×0.02500 dm3\text{Moles of HCl} = 0.100 \, \text{mol dm}^{-3} \times 0.02500 \, \text{dm}^3 Moles of HCl=0.002500 mol\text{Moles of HCl} = 0.002500 \, \text{mol}
  7. Use the mole ratio to find the moles of NaOH.

    Since the mole ratio of NaOH to HCl is 1:11:1:

    Moles of NaOH=Moles of HCl=0.002500 mol\text{Moles of NaOH} = \text{Moles of HCl} = 0.002500 \, \text{mol}
  8. Calculate the concentration of NaOH.

    CNaOH=Moles of NaOHVNaOHC_{\text{NaOH}} = \frac{\text{Moles of NaOH}}{V_{\text{NaOH}}} CNaOH=0.002500 mol0.02005 dm3C_{\text{NaOH}} = \frac{0.002500 \, \text{mol}}{0.02005 \, \text{dm}^3} CNaOH=0.124688... mol dm−3C_{\text{NaOH}} = 0.124688... \, \text{mol dm}^{-3}

    Rounding to three significant figures:

    CNaOH=0.125 mol dm−3C_{\text{NaOH}} = \boxed{0.125 \, \text{mol dm}^{-3}}

    (Note: This calculation uses hypothetical values. You would substitute your actual average volume of NaOH and the known concentration and volume of the acid used.)

    • c) State one possible source of error in the experiment and how it could be minimized.

    One possible source of error is overshooting the endpoint during titration. This occurs when too much of the titrant (e.g., NaOH) is added beyond the point where the reaction is complete, leading to an inaccurate volume reading. This error can be minimized by adding the titrant drop by drop as the endpoint is approached, especially after the indicator starts to show temporary color changes, and by swirling the flask continuously to ensure thorough mixing.

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