Question

If normal human plasma contains about 3 mEq/L of hydrogen phosphate ion HPO4⁻², how many mg/L of dibasic potassium phosphate K₂HPO₄ (MW 174), required to supply the needed HPO4⁻² for an electrolyte replacement in the hospital?

Answer 1

Approximately 522 mg/L of dibasic potassium phosphate (K₂HPO₄) is required to supply the needed HPO₄²⁻ for an electrolyte replacement in the hospital.

Step-by-step explanation:

To calculate the amount of dibasic potassium phosphate (K₂HPO₄) needed to supply the required hydrogen phosphate ion (HPO₄²⁻), we can use the equation:

HPO₄²⁻ + H₂O → H₂PO₄⁻ + OH⁻

The mol ratio between HPO₄²⁻ and K₂HPO₄ is 1:1, which means that for every 1 mol of HPO₄²⁻, we need 1 mol of K₂HPO₄. We can use the molar mass of K₂HPO₄ (174 g/mol) to convert from moles to grams. Finally, we can use the given concentration of 3 mEq/L of HPO₄²⁻ to calculate the required mass of K₂HPO₄ in mg.

Let's calculate:

1. Convert 3 mEq/L of HPO₄²⁻ to moles by dividing by 1000: 3 mEq/L ÷ 1000 = 0.003 mol/L.
2. Multiply the mol ratio by the number of moles of HPO₄²⁻: 0.003 mol/L × 1 mol K₂HPO₄/1 mol HPO₄²⁻ = 0.003 mol/L.
3. Convert the mol/L to mg/L by multiplying by the molar mass of K₂HPO₄: 0.003 mol/L × 174 g/mol × 1000 mg/g = 522 mg/L.

Therefore, approximately 522 mg/L of dibasic potassium phosphate (K₂HPO₄) is required to supply the needed HPO₄²⁻ for an electrolyte replacement in the hospital.