Question

Define the terms: (a) Salinity; (b) Ionic Strength; and calculate their values for a water body of pH 9 and density 1 kg/L, with ionic content: 5.55 x 10-5 M Calcium; 1.29 x 10-4 M Chloride; 1.76 x 10-5 M Potassium; 4.45 x 10-5 M Magnesium; 3.22 x 10-5 M Sulfate; and 1.36 x 10-4 M Bicarbonate.

Answer 1

Salinity : The no of grams of salts dissolved in one liter of water. (g/L)

Ionic strength : Half of the sum of concentrations of all ions present in water multiplied with their charges.

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The water with pH= 9 is basic means contains OH- ions too

pOH = 14-pH = 14-9 = 5

pOH = 5 = -log[OH-]

antilog of -5 = [OH] = 1 x 10^-5

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No of grams of Ions = Molarity of Ions x Ionic weight

For OH- ions

= 1 x 10^-5 x 17 = 1.7 x 10^-4 g

For Ca²⁺ ions

= 5.55 x 10^-5 x 40 = 22.2 x 10^-4 g

For Cl^-ions

= 1.29 x 10^-4 x 35 = 45.15 x 10^-4 g

For Mg²⁺ ions

= 4.45 x 10^-5 x 24 = 10.7 x 10^-4 g

For SO₄^2- ions

= 3.22 x 10^-5 x 96 = 31 x 10^-4 g

For HCO₃^- ions

= 1.36 x 10^-4 x 62 = 84 x 10^-4 g

Salinity = sum of masses of all ions

= 1.94 x 10^-2 grams

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IonicstrengtC1.21+C2.222 +

= 1/2 (1 x 10^-5 x 1²)+(5.55 x 10^-5 x 2²) + (1.29 x 10^-4 x 1²) + (4.45 x 10^-5 x 2²) + (3.22 x 10^-5 x 2²) + (1.36 x 10^-5 x 1²)

= 1/2 (1 x 10^-5 + 22.2 x 10^-5 + 12.9 x 10^-5 + 17.8 x 10^-5 + 12.88 x 10^-5 + 1.36 x 10^-5 )

= 1/2 (86 x 10^-5)

= 4.3 x 10^-4