Question

Pre-Laboratory Assignment 1. Briefly describe the hazards associated with the Ca2+ ion-precipitating buffer solution you will use in this experiment (in Part I). enim o larak 2. Briefly explain what is meant by the statement, "Ca2+ ion is quantitatively removed from solution by precipitation with C20.-ion." 3. Briefly distinguish between the terms temporary water hardness and permanent water hardness as they pertain to this experiment.pato 4. Suppose that a liter of sample water contains 1.00 mmol of Ca2+ ions and 1.50 mmol of HCO3-ions. (1) How many millimoles of CaCO, will precipitate if this water is boiled for 15 min? (2) How many millimoles of Cations will remain in solution after the water has been boiled?

questions 2-4(1,2)

Answer 1

Here we identify the ca 2+ and modt ion Concentration Water Sample we can in the do this by quantitatively ppt ca at ion as calcium oxalate (cac204) from a given known volume of Water, using cat ion-precipitation buffer solution. The buffer soteliom compelse (Hy) 2 5204. Cammonium oxalate), NH4U Cammonium Chloride) NH404 ammonium hydroride Below is the equation: - Ca2+ (aq) + C2042 (aq) (ac 204 Solid, white) Calzoy is very insoluble in water, the addition of eness of onalak lon (c2oy2-) will precipitate the manimum Quantity of cat in possible so that it willl not interfere with the next tibration. We remove the precipitate by filferation, then analyze the filterate par mqdt ion by threation with EDTA solution. The dessimilarity bet calculated total hardness and Mg 2+ ion hardness is the cakt in hardness in the water sample 16 Tempeoray hardness - It alludes to the hardness that is - Caused by the existence of bicarbonates of Mg and ca. that can be simply removed by boiling the water hence the term we use is temporary Permanent handness - The hardness that cannot be - removed by boiling, because it is caused by the presence of sulfat, salt of ce, that is e not precipitated

@ @ [call ] = moot/L= 0.001M . - Iltacing 8] = 115 mmol/l = 070015M Ksp For Calog = 3.36X109 K&P= [с'][eos 3136X169 = (n)(a) as 5:80810*5 m molay Solubility of Calog 3:36 X1077 / 5480 X 10 -00000 57 93103 5.79x109 mmol predpitald The Amount of Ca Remaining in the Solution 5:19X103 -0.00114 03 X/03 =19.4 X1079