posted by Anonymous on .
The concentration of calcium ions in hard water can be determined by titration with reagent X,which forms a complex with Ca2+(aq), giving a change of color. Three moles of X combine with one mole of Ca2+(aq). Since the color change in this titration is sensitive to pH, the titration has to be carried out in an alkaline buffer. A 25cm3 sample of hard water reacted with 24.0 cm3 of 1.00 x 10-4cm3 X(aq)
(a) What is the concentration, in moldm-3, of calcium ions in the hard water?
(b)The structure of X is similar to that of an amino acid. The way X acts as a ligand can be understood by considering how aminoethanoic acid, H2NCH2CO2H, forms coordinate(dative covalent) bonds with cations
(c)Explain why the complex formed betwwen aminoethanoic acid and aqueous calcium ions is more stable at pH10 than at pH4
I think you have a typo here. I believe the 1.00 x 10^-4 cc X should be 1.00 x 10^-4 mol/dm^3 of X.
24.0 cc X * 1.00E-4 mol/dm^3 = 0.0024 mmols X.
Divide by 3 to obtain mmols Ca^2+ = 0.0008.
a. (Ca^2+) = mmols/cc.
b. no question.