FE Other Disciplines Chemistry Practice Problems

The current FE Other Disciplines specification assigns 5-8 of the 110 exam questions to Chemistry. Form A contains 6 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A groundwater supply contains 96 mg/L of magnesium ion ( $Mg^{2 +}$ ). Molar masses are $Mg = 24.31$ g/mol and $CaCO_{3} = 100.09$ g/mol. The magnesium hardness expressed as mg/L as $CaCO_{3}$ is most nearly:

(A)46.6 mg/L as $CaCO_{3}$
(B)198 mg/L as $CaCO_{3}$
(C)240 mg/L as $CaCO_{3}$
(D)395 mg/L as $CaCO_{3}$

Show worked solution

Answer: (D)

Reporting the divalent magnesium on an equivalent basis yields about 395 mg/L as calcium carbonate.

Hardness reporting restates every polyvalent ion as the chemically equivalent amount of calcium carbonate, so the ion concentration must first be expressed in equivalents using its charge of 2.

E W_{Mg} = \frac{24.31}{2} = 12.16 g/eq

\frac{96}{12.16} = 7.90 meq/L

That same equivalent count is then priced at the equivalent weight of calcium carbonate, itself a divalent salt.

E W_{CaCO_{3}} = \frac{100.09}{2} = 50.05 g/eq

C = 7.90 (50.05) = 395 mg/L as CaCO_{3}

FE Reference Handbook — Chemistry: Equivalent Weight and Normality (Hardness as CaCO3)

Why the other choices appear

(A)Inverts the conversion factor, multiplying the ion concentration by $24.31/50.05$ to obtain 46.6 mg/L.
(B)Mismatches the weights by dividing by the magnesium molar mass while multiplying by the carbonate equivalent weight: $(96/24.31) (50.05) = 198$ mg/L.
(C)Uses the equivalent weight of calcium (20.04 g/eq) in place of magnesium's: $(96/20.04) (50.05) = 240$ mg/L.