Question

given an aqueous Solution that contains 150 mg/L Ca^+2, 160 mg/L Mg^+2, 100 x 10^(-6) g/L...

given an aqueous Solution that contains 150 mg/L Ca^+2, 160 mg/L Mg^+2, 100 x 10^(-6) g/L OH^-, and 120mg/L HCO3^-.

A) what is the Total Hardness in mg / L as CaCO3?

B) pH ?

C) Will bicarbonate Concentration (HCO3^-) accurately represent alkalinity (alk) ? why or Why Not ?

Homework Answers

Answer #1

[Ca^2+] = 150mg/L *50/(40/2) = 375 mg/L as CaCO3

[Mg^2+] = 160mg *50/(24/2) = 666.67 mg/L as caCO3

[HCO3^-] = 120mg/L *50/61/1 = 98.36 mg/L

total hardness = 375 + 666.67+ 98.6 = 1140.27 mg/L

----------------------------

pOH = -log[OH-] = -log[100*10^-6] = 4

pH = 14-pOH = 14-4 = 10

---------------------------------------

Alkalinity includes the concentration of carbonate , concentration of H+, OH- and HCO3-. So, concentration of HCO3- alone does not accurately represent alkalinity.

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