Many common cations, such as Na+ and K+, are not readily analyzed by simple titrametric or gravimetric procedures. These cations can be analyzed quantitatively by the technique of ion exchange. In this case the Na+ and K+ are titrated with a standardized strong base solution. The reactions involved are as follows:
Ion exchange reaction
Na+ + H-(Resin) → Na-(Resin) + H+
H+ + OH- → H2O
If cations such as Ca2+ are present, they also will react with the resin to liberate H+ and the proton will be titrated in the titration reaction. Furthermore, any strong acid in the original solution will pass through the resin unchanged and can be titrated. To take care of a problem such as this, a person has to resort to the use of multiple techniques to analyze the system.
Consider starting with a solution containing H+, Ca2+, and Na+. H+ can be titrated with strong base and quantified directly. Ca2+ can be titrated with ethylenediaminetetraacetic acid (EDTA) and quantified directly. Na+, however, can only be quantified indirectly. By exchanging both the Na+ and Ca2+ for H+ on the ion exchange column, the amount of Na+ in the original sample can be determined using a difference calculation (total H+ passed through the column minus the amount of H+ in the original solution, minus twice the amount of Ca2+ in the original solution since each Ca2+ ion liberates 2 H+ ions).
If an aliquot of the original sample contains X mmoles of Ca2+, Y mmoles of Na+, and Z mmoles of H+, one can see that after the ion exchange step there are (2X + Y + Z) mmoles of strong acid to be titrated with the strong base. Titration of another aliquot of the sample with EDTA will allow one to find the number of mmoles of Ca2+ (X) in the sample. Titration of still a third aliquot with the standardized strong base will give the number of mmoles of strong acid (Z) in the sample. The number of mmoles of Na+ (Y) is found from these data using the equation.
Aliquot: 10 mL of a 250 mL diluted sample.
50 mL of H20 were added to the aliquot and then titrations were started.
Titrations using NaOH as titrant: 19.55 mL, 18.65 mL, 18.70 mL NaOH standardization was= 0.083989 M
Titrations using EDTA as titrant: 26.65 mL, 24.65 mL, 24.80 mL EDTA Standardization was= 0.01 F EDTA
Titrations through the ion exchange column and EDTA as titrant: 33.75 mL, 33.45 mL, 33.20 mL
Calculate the concentrations of H+, Na+, and Ca2+ in units of N, µg/mL, and ppm CaCO3 respectively.
to find the concentration of Ca2+ ions we need to use formula N1V1=N2V2
N1= normality of Ca2+
N2= normality of EDTA
so, 2 * N1 * *1000*60/250 = 2* (26.65+24.65+24.80)/3*0.01
this gives N1 = 0.105*10-2 M = 1.05 µg/mL, 105 ppm
similarly concentration of H+ ions will be
N3= 0.66*10-2M , 6.663 µg/mL , 666 ppm
total moles of ions used will be
N4 = ((33.75+33.45+33.20)/3)*0.01*250/60000
N4= 0.001366 M
now we know that concentration of Na+ ion will be
N5 = 0.001366-(2*0.001051)+0.00663 = 0.007083 M = 0.783 µg/mL, 783 ppm
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