Question

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9. (13 points) A sample of water has the following concentrations of ions, assuming no other constituents are missing: Cations mg/L Anions mg/L Ca 40 HCOị 100 Mg? 10 CO;2- 100 Na ? SO/2 67.2 K Ch 11 н — | ОН ? (a) If the solution has the pH of 10.2, what are the concentrations of H' and OH? For H,O-> H + OH, the equilibrium constant equals 10.(3 points) (b) Use an anion-cation balance to estimate the concentration of Na. (3 points) (c) What is the total hardness (TH)? (3 points) What's the TDS? (2 points) What's the alkalinity? (2 points)

Answer 1

9) a) H₂O -> H⁺ + OH^- Equilibrium Constant = 10^-14

In water, [H⁺] = [OH^-] = 10^-7

pH = -log [H⁺] = 10.2

[H⁺] = 6.3096 x 10^-11

Total [H⁺] = 6.3096 x 10^-11 + 1 x 10^-7

= 1.00063 x 10^-7 mol/L

Total [H⁺] = 1.00063 x 10^-4 mg/L

pOH = 14 - pH = 3.8

[OH^-] = 1.58489 x 10^-4

Total [OH^-] = 1.58489 x 10^-4 + 1 x 10^-7

= 1.58489 x 10^-4 mol/L

= 1.58489 x 10^-4 mol/L x 17 g/mol

Total [OH^-] = 2.69 mg/L

b) Based on anion-cation balance

According to principles of electroneutrality, the total charge of the aqueous solution is zero.

sum of no. of equivalents of cations = sum of no. of equivalents of anion

In the table, given that mass of ions in 1L.

No. of equivalents of Ca²⁺ = 40 mg/L / 40 g/mol x 2 = 2 milliequivalents/L

No. of equivalents of Mg²⁺ = 10 mg/L / 24.305 g/mol x 2 = 0.822 milliequivalents/L

No. of equivalents of K⁺ = 7 mg/L / 39.0983 g/mol x 1 = 0.179 milliequivalents/L

No. of equivalents of H⁺ = 1.00063 x 10^-4 mg/L / 1 g/mol x 1 = 0 milliequivalent/L

No. of equivalents of HCO₃^- = 100 mg/L / 61.0168 g/mol x 1 = 1.6 milliequivalents/L

No. of equivalents of CO₃^2- = 100 mg/L / 60.009 g/mol x 2 = 3.33 milliequivalents/L

No. of equivalents of SO₄^2- = 67.2mg/L / 96.06 g/mol x 2 = 1.399 milliequivalents/L

No. of equivalents of Cl^- = 11 mg/L / 35.453 g/mol x 1 = 0.31 milliequivalents/L

No. of equivalents of OH^- = 2.69 mg/L / 17.008 g/mol x 1 = 0.158 milliequivalents/L

sum of no. of equivalents of cations = sum of no. of equivalents of anion

No. of equivalents of Na⁺ = Sum of no. of equivalents of anion - sum of no. of equivalents of cation except sodium ion

= 6.797 - 3.001 = 3.796 milliequivalents/L

Concentration of Na⁺ = (3.796/1)mmol/L x 22.9897 g/mol

= 87.2689 mg/L

c) Total Hardness is the measure of the amount of dissolved calcium and magnesium ions in water.

d) TDS: Total dissolved Solid is the measure of total concentration of the dissolved inorganic salts, heavy metals as well as some amount of organic matter in water.

e) Alkalinity is the measure of the amount of alkaline substances present in the water.