Question

You are determining the water hardness of sample from a VCU water fountain. You add 30.00 mL of the water sample to a 500. mL Erlenmeyer flask and all the other solutions needed to perform the titration. The sample requires 16.53 mL of a 3.365×10^-3 M EDTA titrant solution to reach the endpoint. The blank requires 0.05 mL of EDTA. What is the water hardness of the sample in units of (moles Ca2+ / L)? Use the assumptions provided in the Data Analysis section of the handout.

Answer 1

Answer:

Volume of water sample = 30 mL (given)

Concentration of EDTA = 3.365 x 10^-3 M (given)

                              = 3.365 x 10^-3 moles / L         [ since 1 M = 1 moles / L ]

                              = 3.365 x 10^-3 x 1000 milli moles / 1000 mL [ because 1 mole = 10³ milli moles and 1 L = 10³ mL]

                             = 3.365 x 10^-3 millimoles / mL [ thus 1 moles / L = 1 millimoles / mL]

Titre value = 16.53 mL (given)

Blank value = 0.05 mL (given)

Therefore Titre value after blank correction = 16.48 mL     [ = 16.53 - 0.05]

millimoles of EDTA used for reaction with the hardness in the sample = 16.48 mL * 3.365 * 10^-3 milli moles / mL

                                                                                                     = 5.5455 x 10^-2 millimoles.

one millimole of EDTA is combines with 1 millimole Ca²⁺ ion to form Ca-EDTA complex. as per the equation

             Ca²⁺ + EDTA^4-         [Ca(EDTA)]^2-

Therefore 5,5455 x 10^-2 millimoles of EDTA is combines with 5.5455 x 10^-2 millimoles of Ca²⁺

Therefore millimoles of Ca²⁺ present in the given sample of 30 mL is 5.5455 x 10^-2

Hence the concentration of Ca²⁺ is = 5.5455 x 10^-2 millimoles / 30 mL

                                                     = 1.8485 x 10^-3 millimoles / mL

                                                     = 1.85 x 10^-3 moles / L           [ since 1 millimoles / mL - 1 moles / L]

Therefore water hardness of the sample as Ca²⁺ = 1.85 x 10^-3 moles/L