Due December 3rd The TAP water and DI water samples were both titrated with 0.00

Question

Due December 3rd

The TAP water and DI water samples were both titrated with 0.005 M EDTA.Sample

TAP Water

DI Water

Volume of 0.005 M EDTA used 150 mL

75 mL

Analysis: Using the data above, calculate the concentration of metal ions in both the TAP and DI water
samples. What is the concentration (mol/L) of hard water ions ALREADY present in the lab’s tap water?Based on the information, would you recommend that the university purchase water softeners? (Hint:you should look up the hard water concentration limit suggested by the EPA.)
With this information, write a formal lab report. Please refer to the formal lab report gradingrubric as well as the formal lab report guidelines.
Please contact your TA if you have any questions regarding this lab. This report must be submittedthrough Turnitin and a hard copy must be handed in on December 3rd.

Explanation / Answer

The hardness of water is due to the presence of calcium and magnesium ions

1. Determination of the total concentration of calcium and magnesium ions in the lab water sample.

concentration of EDTA = 0.005 mol L-1
   volume of EDTA solution = 150mL

Volume of water sample = 1000mL

moles EDTA = concentration (mol L-1) x volume (L) = 0.005 x 150mL = 0.75 mol

2. Write the equation for the reaction between EDTA and Ca2+ and Mg2+:
Ca2+ + H2EDTA2- CaEDTA2- + 2H+
Mg2+ + H2EDTA2- MgEDTA2- + 2H+
Let M2+ = Ca2+ + Mg2+, then
M2+ + H2EDTA2- MEDTA2- + 2H+

3. Calculatation of moles of M2+ (= Ca2+ + Mg2+) present in the 1000 mL water sample
From the balanced chemical equation: 1 mole M2+ reacts with 1 mole EDTA
So 0.75 moles of EDTA react with 0.75moles of M2+

4. Calculation of concentration of M2+ (= Ca2+ + Mg2+) in lab water sample
concentration (M2+) = moles M2+ ÷ volume (L)
moles M2+ = 0.75 mol
volume = 1000 mL =1L
concentration (M2+) = 0.75 ÷ 1 L= 0.75 mol L-1

5.The total concentration of dissolved calcium ions and magnesium ions in the lab water is 0.75 mol L-1

The hardness of water is due to the presence of calcium and magnesium ions

1. Determination of the total concentration of calcium and magnesium ions in the lab water sample.

concentration of EDTA = 0.005 mol L-1
   volume of EDTA solution = 150mL

Volume of water sample = 1000mL

moles EDTA = concentration (mol L-1) x volume (L) = 0.005 x 150mL = 0.75 mol

2. Write the equation for the reaction between EDTA and Ca2+ and Mg2+:
Ca2+ + H2EDTA2- CaEDTA2- + 2H+
Mg2+ + H2EDTA2- MgEDTA2- + 2H+
Let M2+ = Ca2+ + Mg2+, then
M2+ + H2EDTA2- MEDTA2- + 2H+

3. Calculatation of moles of M2+ (= Ca2+ + Mg2+) present in the 1000 mL water sample
From the balanced chemical equation: 1 mole M2+ reacts with 1 mole EDTA
So 0.75 moles of EDTA react with 0.75moles of M2+

4. Calculation of concentration of M2+ (= Ca2+ + Mg2+) in lab water sample
concentration (M2+) = moles M2+ ÷ volume (L)
moles M2+ = 0.75 mol
volume = 1000 mL =1L
concentration (M2+) = 0.75 ÷ 1 L= 0.75 mol L-1

5.The total concentration of dissolved calcium ions and magnesium ions in the lab water is 0.75 mol L-1

The hardness of water is due to the presence of calcium and magnesium ions

1. Determination of the total concentration of calcium and magnesium ions in the DI water sample.

concentration of EDTA = 0.005 mol L-1
   volume of EDTA solution = 75mL

Volume of water sample = 1000mL

moles EDTA = concentration (mol L-1) x volume (L) = 0.005 x 75mL = 0.375 mol

2. Calculatation of moles of M2+ (= Ca2+ + Mg2+) present in the 1000 mL water sample
From the balanced chemical equation: 1 mole M2+ reacts with 1 mole EDTA
So 0.375 moles of EDTA react with 0.375moles of M2+

3. Calculation of concentration of M2+ (= Ca2+ + Mg2+) in DI water sample
concentration (M2+) = moles M2+ ÷ volume (L)
moles M2+ = 0.375 mol
volume = 1000 mL =1L
concentration (M2+) = 0.375 ÷ 1 L= 0.375 mol L-1

4.The total concentration of dissolved calcium ions and magnesium ions in the DI water is 0.375 mol L-1

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