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^{-} → H_{2}O

If cations such as Ca^{2+} 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^{+},
Ca^{2+}, and Na^{+}. H^{+} can be titrated
with strong base and quantified directly. Ca^{2+} can be
titrated with ethylenediaminetetraacetic acid (EDTA) and quantified
directly. Na^{+}, however, can only be quantified
indirectly. By exchanging both the Na^{+} and
Ca^{2+} 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 Ca^{2+} in the original
solution since each Ca^{2+} ion liberates 2 H^{+}
ions).

If an aliquot of the original sample contains *X* mmoles
of Ca^{2+}, *Y* mmoles of Na^{+}, and
*Z* mmoles of H^{+}, one can see that after the ion
exchange step there are (2*X* + *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 Ca^{2+} (*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 Ca^{2+} in units of N, µg/mL, and ppm
CaCO_{3} respectively. **

Answer #1

to find the concentration of Ca^{2+} ions we need to use
formula N1V1=N2V2

N1= normality of Ca^{2+}

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^{-2}M , 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**

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