Question

Water Hardness by EDTA Titration
Objective: To determine [Ca2+,MG2+] and [Mg2+] concentration in an unknown water sample

A 0.0107 molarity solution of EDTA is prepared for the experiment.

25 mL of an unknown is pipetted and diluted 100 mL with distilled water and adjusted to a pH between 8 and 10 with NH4OH. A 10.00 mL portion of the diluted unknown is transferred to a 250 mL flask with 3 mL of pH 10 ammonia buffer and tiny amount of Erichrome Black T indicator (not diluted with distilled water).

The unknown is titrated with EDTA to an endpoint:

Trial 1: 18.26 mL
Trial 2: 17.66
Trial 3: 17.75
Trial 4: 17.73

A 10.00 mL aliquot of the unknown id pipetted into an Erlenmeyer flasks and 50 mL of distilled water is added to the flask. 10 drops of 50% KOH is added to precipitate Mg (OH)2 while remaining at a pH between 12 and 12.5. EDTA is titrated into the sample for three different trials:

Trial 1: 9.88 mL
Trial 2: 9.78 mL
Trial 3: 9.32 mL

The first set of trials is to calculate the [Ca2+, Mg2+] while the second set of trials is to calculate [Ca2+] alone. How are these values calculated??

Answer 1

Molarity of EDTA solution = 0.0107M

Now the unknown solution has both [Ca2+] and [Mg2+]. Let this net concentration be X.

Metal ions form 1:1 complex with EDTA

1 mole of EDTA = 1 mole of net metal ion

1000mL of EDTA solution has = 0.0107 moles of EDTA

Trial1: 18.26 mL of EDTA solution has = 0.0107*18.26/1000 = 1.953*10^-4 moles

Trial 2: 17.66 mL of EDTA solution has = 0.0107*17.66/1000 = 1.889*10^-4 moles

Trial 3: 17.75 mL of EDTA solution has = 0.0107*17.75/1000 = 1.899*10^-4 moles

Trail 4: 17.73 mL of EDTA solution has = 0.0107*17.73/1000 = 1.897*10^-4 moles

Thus net concentration of Ca2+ and Mg2+ is 10ml aliquot of diluted solution is equal to moles of EDTA consumed.

Mean concentration = 1.9095*10^-4 moles of Ca2+ and Mg2+ in 10ml aliquot of diluted.

For estimation of Ca2+ alone,

Moles of EDTA consumed = Moles of Ca2+

Trial 1: 9.88 mL of EDTA has = 0.0107*9.88/1000 = 1.057*10^-4 moles

Trial 2: 9.78 mL of EDTA has = 0.0107*9.78/1000 = 1.046*10^-4 moles

Trial 3: 9.32 mL of EDTA has = 0.0107*9.32/1000 = 0.997*10^-4 moles

Hence moles of Ca2+ present in 10mL aliquot of diluted unknown = moles of EDTA consumed above.

Thus mean concentration of Ca2+ = (1.057*10^-4 + 1.046*10^-4 + 0.997*10^-4 )/3 = 1.033*10^-4 moles of Ca2+

Hence moles of Mg2+ = (Total Mg2+ and Ca2+ ) - (Moles of Ca2+)

= 1.9095*10^-4 - 1.033*10^-4 moles = 0.8762*10^-4 moles of Mg2+

thus 10 ml aliquot has 0.8762*10^-4 moles of Mg2+ and 1.033*10^-4 moles of Ca2+

100 mL of diluted solution has = 0.8762*10^-4 *100/10moles of Mg2+ and 1.033*10^-4 *100/10moles of Ca2+

= 0.8762*10^-3 moles of Mg2+ and 1.033*10^-3 moles of Ca2+

Original concentration of Mg2+ and Ca2+ in 25 ml of undiluted solution

= 0.8762*10^-3 moles of Mg2+ and 1.033*10^-3 moles of Ca2+ in 25 mL of undiluted stock (no of moles doesnt change on dilution)

= 0.8762*10^-3 *1000/25 moles of Mg2+ and 1.033*10^-3 *1000/25 moles of Ca2+ in 25 mL of undiluted stock

= 0.035 M of Mg2+ and 0.04132 M of Ca2+

Hence the undiluted stock = 0.035M in Mg2+ and 0.04132M in Ca2+