Question

RAW DATA AND OBSERVATIONS PART 1 DETERMINATION OF TOTAL HARDNESS Concentration of standardized EDTA(ag) O.06162 Table 1-Titrations of 25.00 mL aliquots of mineral water with -0.01 mol L EDTA(aq at pH 10 using Calmagite® as indicator (2,5 marks) Sample 2 l 3 4 Fmol bue He i3 96 b.05 21-70 Volume CmL) Inital bunette 3 46 5.60 i7-30 volume Cm) 10 46 Volume of EDTA oddud Cmb iO.40

Answer 1

ml. = 10.45 m. we know that, v. xS. = V₂ X 2 1. = Average volume of EDTA = 10.45 + 10.50 + 10.40 m. Si = Strength of EDTA = 0.05152 mol/l V2 = Volume of water sample = 25 ml. de = ? So, 10.45 X 0.05152 = 25x52 S2 = 18.45 X0.05152 melse = 0.02153 mill - 25 So, the concentration of catt/eyet in the mineral water = 0.02153 @ 1 mole calo = 100 gm caloz = 100000 my caco, In part 1, we have found that the water sample contains 0.02153 mol/L caction = (0.02153 x 100000) my /L CaCO₃ = 2153 mg caloz in 1L water Therefore the concentration is 2153 ppm

The concentration of metal ion in the mineral water can be calculated using the formula, V₁ X S₁ = V₂ X S_2.

V_1. = Volume of EDTA = 10.45 ml

S₁ = Strength of EDTA = 0.05152 mol / L

V₂ = Volume of water = 25 ml

S₂ = V₁S₁/V₂

From here, the strength of water in terms of the concentration of calcium / magnesium ion can be calculated in mol / L unit.

It can further be converted into ppm unit by multiplying with the molecular weight of calcium carbonate.