Question

A water analysis by a commerical analytical labortatory listed the following results of primary cations and anions in a submitted water sample.

Cations - Ca2+ (93.8 mg/L) , Mg2+ (28.0 mg/L) , Na+ (13.7 mg/L) , K+ (30.2 mg/L)

Anions - HCO- (164.7 mg/L) , SO4 2- (134.0 mg/L) , Cl- (92.5 mg/L)

Do these results violate the principle of electroneutrality? Acceptable error - 0.1065 + (0.0155 x sum of anions)

Provide a likely reason that H+ and OH- were not listed in these results.

What is the hardness of this water?

Answer 1

Calculations for hardness of water -

Constituents reported...........Concentrations.............Equivalents of CaCO3

Ca 2+.........................................93.8mg /L................[ ( 93.8 x 100 ) / 40 ] =234.5 mg / L

Mg 2+........................................28.0 mg / L................[ ( 28 x 100 ) / 24 ] = 116.6 mg / L

Na +......................................... 13.7 mg / L .......... .[ ( 13.7 x 100 ) / 2 x 23] = 29.78 mg/ L :

(ignored as it does not contribute to hardness)........................................

K +...........................................20.2 mg / L ...............[ ( 30.2 x 100 ) / 2 x 39 ] = 38.72 mg / L .............................. .....

( ignored as it does not contribute to hardness ).........................................

HCO3 -.....................................164.7 mg / L ..............[ (164.7 x 100 ) / 2 x 61 ] = 135.0 mg /L

SO4 2-..................................... 134.0 mg / L....... .....[ (134 x 100 ) /96=139.58.mg/L................................................... ( ignored as it does not contribute to hardness) .................................

Cl - ......................................... 92.5 mg / L ................[ (92.5 x!00 )/71 ] = 130.28 mg /L .................................ignored as it does not contribute to hardness.................................

Total hardness = ( 234.5 + 116.6 + 135 .0 ) mgm / L

.............................= 486 .16 mg / L

........................or = 486 .16 ppm

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Eventhough the ions (viz. K+ , Na+ , Cl-, SO4 2- do not contribute to hardness, their equivalents in terms of CaCO3 are required to be calculated to decide about the electroneutrality The same way as above . Thus,

the amounts of cations and anions are found to be approximately equal. This results into maintanance of electrical neutrality.

Hence , it should not be necessary to list H+ & OH- in the result.