Question

The US Environmental Protection Agency (EPA) places limits on the quantities of toxic substances that may be discharged into the sewer system. Limits have been established for a variety of substances including hexavalent chromium which is limited to 0.50 mg/L .If an industry is discharging hexavalent chromium as potassium dichromate (K2Cr2O7), what is the maximum permissible molarity of that substance?

Answer 1

The maximum permissible molarity of potassium dichromate (K₂Cr₂O₇) in wastewater, as regulated by the EPA for hexavalent chromium (Cr(VI)), is 3.395 × 10⁻⁶ M, based on the set limit of 0.50 mg/L for Cr(VI).

Step-by-step explanation:

To calculate the maximum permissible molarity of potassium dichromate (K₂Cr₂O₇) in wastewater, according to the US Environmental Protection Agency (EPA) regulations, one must consider the molar mass of K₂Cr₂O₇ and the limit set for hexavalent chromium (Cr(VI)), which is 0.50 mg/L.

The molar mass of K₂Cr₂O₇ is 294.185 g/mol. Since every mole of potassium dichromate contains two moles of Cr(VI), we calculate the maximum molarity by dividing the EPA limit for Cr(VI) by the molar mass of potassium dichromate, then multiplying by 2 to account for the stoichiometry:

Molarity (M) = (mg/L ÷ (g/mol)) × (1 mol / 1000 mg) × 2

Molarity (M) = (0.50 mg/L ÷ 294.185 g/mol) × (1 mol / 1000 mg) × 2 = 3.395 × 10⁻⁶ M

Therefore, the maximum permissible molarity of potassium dichromate in the wastewater is 3.395 × 10⁻⁶ M.

Answer 2

4.8x10⁻⁶ M

Step-by-step explanation:

The question asks for the maximum molarity of K₂Cr₂O₇ that can be discharge into the sewer system if the limit of hexavalent chromium (Cr(VI)) that is permmitted to discharge is 0.50 mg/L.

Molarity is defined as moles of K₂Cr₂O₇ per Liter, so we need to calculate to how many moles of K₂Cr₂O₇ corresponds the 0.50 mg of Cr(VI).

In 1 mol of K₂Cr₂O₇ there are 2 moles of Cr(VI) which are 103.99 g of Cr(VI) (2 * molar mass of Cr = 2 * 51.9961 g/mol)

So 103.99 g of Cr(VI) can be found in 1 mol of K₂Cr₂O₇

We need to calculate in how many moles of K₂Cr₂O₇ we can find 0.50 mg or 5.0x10⁻⁴ g of Cr(VI) (1 g = 1000 mg) :

x moles K₂Cr₂O₇ / 5.0x10⁻⁴ g Cr(VI) = 1 mol K₂Cr₂O₇ / 103.99 g Cr(VI)

x = 5.0x10⁻⁴ g * (1 mol K₂Cr₂O₇ / 103.99 g Cr(VI)) = 4.8x10⁻⁶ moles K₂Cr₂O₇

4.8x10⁻⁶ moles K₂Cr₂O₇ can be discharged per Liter

So the maximum permissible molarity of K₂Cr₂O₇ is 4.8x10⁻⁶ M