Question

PRE-LABORATORY QUESTIONS, WATER: DETERMINING THE CONCENTRATION OF CALCIUM IONS IN WATER SAMPLES EXPT. SA HARD Prior to coming to lab, carefully read the entire experimental procedure that follows and write an outline of the procedure in your lab notebook IN YOUR OWN WORDS (do not simply copy). Then, add three tables for your titrations of the blank, tap water, and CaCl samples. Complete these problems and hand in gt the bexinning of the laboratory period. Show all work. EDTA is a common chemical species that can be used to determine water hardness, (i.e. the sum of the number of calcium and magnesium ions in a water sample). EDTA (designated as HY) is also known to react in a 1:1 stoichiometry with most metal ions of +2 charge or greater. The titration reaction with calcium ion would be: Ca2. (aq) + H,Y'. (aq) → CaH2Y (aq) When conducting titrations, a common term that is employed is called concentration, which is often defined as the number of moles of a substance (often called the solute)/ L of a solution. This definition is called Molarity and is given the symbol M. 1. (1.5 point) Calculate the Molarity (M) of an EDTA solution if 3.722 g of Na,EDTA 2HyO are dissolved in 1.0 L of water. (Na EDTA 2H,O, MM-372.24 g/mole). Show your work! /halextylMzmeleste salute--o.olomiles一弍0.010 A .00L 372. 2. (2 points total) A 5.00 mL tap water sample was measured out with a volumetric pipette, and added to a 25 mL. Erlenmyer flask. It was then titrated with a 0.0100 M Na,EDTA 2H,O solution, and found to take 0.635 ml of the EDTA solution to reach the blue endpoint. a) How many moles of the 0.0100 M Na EDTA 2H,O were added in the titration reaction? Show calculation. b) How many moles of calcium ions were reacted? Show calculation. 63

Answer 1

2a) Moles of Na₂EDTA.2H₂O = (volume of Na₂EDTA.2H₂O in L)*(concentration of Na₂EDTA.2H₂O) = (0.635 mL)*(1 L/1000 mL)*(0.0100 M) = 6.35*10^-6 mole (ans).

b) Write the balanced chemical equation for the reaction between Ca²⁺ ions and Na₂EDTA.2H₂O as below.

Ca²⁺ (aq) + Na₂EDTA.2H₂O (aq) ---------> CaEDTA (aq) + 2 Na⁺ (aq) + 2 H₂O (l)

As per the stoichiometric equation,

1 mole Na₂EDTA.2H₂O = 1 mole Ca²⁺.

Therefore, mole(s) of Ca²⁺ reacted = 6.35*10^-6 mole (ans).

3) We took 5.00 mL of tap water and the tap water contained 6.35*10^-6 mole Ca²⁺ ions. Therefore, concentration of Ca²⁺ in the tap water = (moles of Ca²⁺)/(volume of solution in L) = (6.35*10^-6 mole)/[(5.00 mL)*(1 L/1000 mL)] = 1.27*10^-3 mol/L = 1.27*10^-3 M (ans).