Question

A titration was performed to standardize an EDTA solution.

a) A 25.00 mL aliquot of a standard solution containing 0.01500 M Ca²⁺ required 42.87 mL of EDTA to reach the endpoint.

The molarity of the EDTA solution is: ________ M

b) That same EDTA solution was then used to titrate an unknown hard water sample. A 40.00 mL aliquot of unknown hard water required 34.21 mL of EDTA solution to reach a distinct endpoint.

The concentration of Ca²⁺ ions (assuming all the hardness is due only to calcium ion) in the solution is: ______ M

Answer 1

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@ The balance equation blw cath and EDTA Ca(EDTA) 2 -2 Cacaq I mot & EDTA cage I mos Here we use MINI = MV formula 12 Cath EDTA M = Molarity of Cat2 sot = 0.0150m) M = Molasity of EDTA sol = ? V = volume of Carth sotn 25.0mL = volume of EDTA2 SO 42.87 mL ni = mole ratio of cat? I mor na mole ratio of EDTA = 1 mol Now M = Movi X 2 = (0.0150M) (25.0ml inds land 42.87mt M = 0.375M = 0.00875 M. 42.87 [EDTA] = 0.00075 M 16 catz + EDTA 5 CA ESTA) ??? M = 0.00875m (from part A answer) va 40.0mL o = 34.21 mL n = 1 mol n₂ = imod > M = ended Marex n1 = (0.00875M) (34.21 ml) mot 40.00 M = 0.299 M = 0.00748M 40.0 d. [cat unknown = 0.0074CM Ca runknown) na