Question

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⁺

            Titration reaction

                                                            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.

Results were:

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.  

Answer 1

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= ((19.55+18.65+18.70)/3)*0.083989*250/60000

N3= 0.66*10^-2M , 6.663 µg/mL , 666 ppm

Similarly,

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