3)A 100 mL sample of hard water is titrated with 22.4 mL of EDTA solution from p

Question

3)A 100 mL sample of hard water is titrated with 22.4 mL of EDTA solution from problem 2. the same amount of MgCl2 is added as previously and the total volume of EDTA solution required is 22.44 mL.
A)What volume of EDTA is used titration the Ca2+ in the hard water?
B)how many moles of EDTA are there in the volume?
C)how many moles of Ca2+ are there in 100 mL of water?
D)assume all of the Ca2+ I'm the water come for CaCo3. how many moles of CaCo3 are in 1L of water?how many grams of CaCo3 are in 1L of water?
E)if 1 ppm CaCo3=1mg/liter, what is the water hardness in ppm CaCO3?

additional info:
1)25.95 mL of EDTA are needed to titrate the Ca2+ ion in the sample.
2)5.12×10^-4 moles of EDTA are in the volume for the above measurement
3)the molarity of the EDTA solution is 1.97×10^-3.

Please use scientific notation if able

Explanation / Answer

When hard water is titrated with EDTA ,both Mg2+ and Ca2+ forms chelating complex of the form MY2- where H2Y2- is the disodium salt of EDTA

Ca2+ (aq)+H2Y2----->CaY2- +2H+

A) if hard water titration uses 22.4ml of EDTA, then it is the volume used up for titration of Mg2+ and Ca2+

if the sample had only Mg2+ ions ,of same molarity then volume of edta used=22.4ml

volume=molarity/moles of edta used up=M/moles

As moles of Ca2+=1/2 the moles of Mg2+ in MgCl2

so volume of edta used for titrating Ca2+=22.4ml /2=12.2 ml

B) molarity of the EDTA solution =1.97*10^-3.M

M(edta)*V(edta)=moles of edta =moles of hard ions(Mg2+and Ca2+) =1.97*10^-3.M*22.4ml*1L/1000ml=0.0004413 moles

C)moles of Ca2+ are there in 100 mL of water=volume of edta needed to titrate Ca2+ * molarity of edta=25.95ml*1.97*10^-3.M=1.97*10^-3.mol/L *0.02595=5.11*10^-4 moles

D)moles of Ca2+=moles of CaCO3=5.11*10^-4 moles

grams of CaCO3 =moles of CaCO3*molar mass of CaCO3=5.11*10^-4 moles*100.0869=0.00511 grams

E) 1 ppm CaCo3=1mg/liter

here concentration of CaCO3=0.00511 grams/100 ml=0.0511g/L=0.0511*1000mg/L=51.1mg/L=51.1 ppm

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