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Math  /  Data & Statistics

Question2. In an experiment to determine the water hardness of tap water, a researcher measured three 50 mL aliquots of water samples and to each sample he added 10 mL of pH 10 ammonia/ammonium chloride buffer and two drops of eriochrome black T indicator and the colour changed to wine red. The resulting solutions were then separately titrated against 0.05 M sodium-EDTA solution from the burette until a final endpoint colour change. The data obtained was recorded in a table as shown below. \begin{tabular}{|l|l|l|l|} \hline Titration & 1 & 2 & 3 \\ \hline final reading /mL/ \mathrm{mL} & 10.70 & 10.65 & 10.60 \\ \hline Initial reading /mL/ \mathrm{mL} & 0.00 & 0.00 & 0.00 \\ \hline Titre value /mL/ \mathrm{mL} & & & \\ \hline \end{tabular} a) What is the expected endpoint colour change? b) Calculate the average titre of the results? c) Write a balance equation of reaction between calcium ions and EDTA d) Calculate the number of moles of EDTA used in each titration and the average number of moles for the whole process. e) What is the concentration of Ca2+\mathrm{Ca}^{2+} (expressed as ppmCaCO3\mathrm{ppm} \mathrm{CaCO}_{3} ) in the sample?

Studdy Solution

STEP 1

1. The initial readings for all titrations are 0.00 mL0.00 \, \text{mL}.
2. The final readings for the titrations are 10.70 mL10.70 \, \text{mL}, 10.65 mL10.65 \, \text{mL}, and 10.60 mL10.60 \, \text{mL}.
3. The concentration of the sodium-EDTA solution is 0.05 M0.05 \, \text{M}.
4. The reaction involves calcium ions and EDTA.

STEP 2

1. Determine the expected endpoint color change.
2. Calculate the average titre value.
3. Write the balanced chemical equation for the reaction.
4. Calculate the moles of EDTA used in each titration and the average moles.
5. Calculate the concentration of Ca2+\mathrm{Ca}^{2+} expressed as ppm CaCO3\mathrm{ppm} \, \mathrm{CaCO}_{3}.

STEP 3

The expected endpoint color change when titrating with EDTA using eriochrome black T indicator is from wine red to blue.

STEP 4

Calculate the titre values:
Titre for 1=10.70 mL−0.00 mL=10.70 mL\text{Titre for 1} = 10.70 \, \text{mL} - 0.00 \, \text{mL} = 10.70 \, \text{mL}
Titre for 2=10.65 mL−0.00 mL=10.65 mL\text{Titre for 2} = 10.65 \, \text{mL} - 0.00 \, \text{mL} = 10.65 \, \text{mL}
Titre for 3=10.60 mL−0.00 mL=10.60 mL\text{Titre for 3} = 10.60 \, \text{mL} - 0.00 \, \text{mL} = 10.60 \, \text{mL}
Calculate the average titre:
Average titre=10.70+10.65+10.603=31.953=10.65 mL\text{Average titre} = \frac{10.70 + 10.65 + 10.60}{3} = \frac{31.95}{3} = 10.65 \, \text{mL}

STEP 5

Write the balanced chemical equation for the reaction between calcium ions and EDTA:
Ca2++EDTA4−→CaEDTA2−\text{Ca}^{2+} + \text{EDTA}^{4-} \rightarrow \text{CaEDTA}^{2-}

STEP 6

Calculate the moles of EDTA used in each titration:
Moles of EDTA for 1=0.05 M×0.01070 L=5.35×10−4 mol\text{Moles of EDTA for 1} = 0.05 \, \text{M} \times 0.01070 \, \text{L} = 5.35 \times 10^{-4} \, \text{mol}
Moles of EDTA for 2=0.05 M×0.01065 L=5.325×10−4 mol\text{Moles of EDTA for 2} = 0.05 \, \text{M} \times 0.01065 \, \text{L} = 5.325 \times 10^{-4} \, \text{mol}
Moles of EDTA for 3=0.05 M×0.01060 L=5.30×10−4 mol\text{Moles of EDTA for 3} = 0.05 \, \text{M} \times 0.01060 \, \text{L} = 5.30 \times 10^{-4} \, \text{mol}
Calculate the average moles of EDTA:
Average moles of EDTA=5.35×10−4+5.325×10−4+5.30×10−43=5.325×10−4 mol\text{Average moles of EDTA} = \frac{5.35 \times 10^{-4} + 5.325 \times 10^{-4} + 5.30 \times 10^{-4}}{3} = 5.325 \times 10^{-4} \, \text{mol}

STEP 7

Calculate the concentration of Ca2+\mathrm{Ca}^{2+} expressed as ppm CaCO3\mathrm{ppm} \, \mathrm{CaCO}_{3}:
The molar mass of CaCO3\mathrm{CaCO}_{3} is 100.09 g/mol100.09 \, \text{g/mol}.
Concentration of Ca2+=5.325×10−4 mol×100.09 g/mol0.050 L=1.065 g/L\text{Concentration of } \mathrm{Ca}^{2+} = \frac{5.325 \times 10^{-4} \, \text{mol} \times 100.09 \, \text{g/mol}}{0.050 \, \text{L}} = 1.065 \, \text{g/L}
Convert to ppm\mathrm{ppm}:
1.065 g/L=1065 ppm1.065 \, \text{g/L} = 1065 \, \text{ppm}
The expected endpoint color change is from wine red to blue. The average titre is 10.65 mL10.65 \, \text{mL}. The balanced equation is Ca2++EDTA4−→CaEDTA2−\text{Ca}^{2+} + \text{EDTA}^{4-} \rightarrow \text{CaEDTA}^{2-}. The average number of moles of EDTA used is 5.325×10−4 mol5.325 \times 10^{-4} \, \text{mol}. The concentration of Ca2+\mathrm{Ca}^{2+} is 1065 ppm CaCO31065 \, \text{ppm} \, \mathrm{CaCO}_{3}.

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