Question

50 mL of 0.010 mol.L^-1 Na₂SO₄   solution was mixed with 50 mL of 0.20 mol.L^-1 KNO₃ solution, calculate the ionic strength of the mixed solution. I need to complete solution, please

Answer 1

Ionic strength of the mixed solution = 0.115

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The ionic strength of the salt solution can be calculated by using the formula,

I = 1/2 Sigma [Ci Zi^2]

I = ionic strength

i = specific ions in the solution

Ci = concentrion of the specfic ions

Zi = Oxidation (valance) of the specific ions, charge of the ion +/- does not consider.

Sigma = Summation of concentrations and valences of all ions in the solution.

Contents of the solution before mixing,

Na2SO4 = 50 mL, 0.010 M

KNO3 = 50 mL, 0.20 M

On mixing these two solution the molarity of the Na2SO4 and KNO3 reduced due to dilution as the volume become 100 mL.

The concentration after mixing,

Na2SO4 = 50 mL x 0.010 M / 100 mL = 0.005 M

KNO3 = 50 mL x 0.20 M / 100 = 0.1 M

Using the formula, I = 1/2 Sigma [Ci Zi^2]

C(Na+) = 0.005 x 2 = 0.01 (1 mole of Na2SO4 produce 2 moles of Na+ ion)

Z(Na+) = 1

C(SO4^2-) = 0.005

Z(SO4^2-) = 2

C(K+) = 0.1

Z(K+) = 1

Z(NO3^-) = 0.1

Z(NO3^-) = 1

I = 1/2 [ (0.01 x 1^2) + (0.005 x 2^2) + (0.1 x 1^2) + (0.1 x 1^2)] = 1/2 [0.01 + 0.02 + 0.1 + 0.1]

I = 1/2 [0.23] = 0.115

Ionic strength = 0.115