Question

A preparation contains, in each milliliter, 236 mg of dibasic potassium phosphate (m.w. = 174.18) and 224 mg of monobasic potassium phosphate (m.w. = 136.09). calculate the total concentration of phosphate, in mmol/ml, and potassium, in meq/ml, in the preparation.6

Answer 1

phosphate = 3.00 mmol/ml potassium = 4.36 meq/ml The formulas for dibasic potassium phosphate and monobasic potassium phosphate are (K2HPO4) and (KH2PO4) respectively. Both will contribute 1 phosphate to the solution and the dibasic will contribute 2 potassium while the monobasic will contribute only 1 potassium. So let's see how many moles of each substance we have. dibasic: 236 mg/ 174.18 g/mol = 1.354920197 mmol monobasic: 224 mg/136.09 g/mol = 1.645969579 mmol Phosphate = 1.354920197 mmol/ml + 1.645969579 mmol/ml = 3.000889776 mmol/ml meq is a rather archaic measurement of concentration standing for milliequivalent. It's a function of the atomic weight of the element and the valence. Since potassium has a valance of 1, the meq/ml value is conveniently equal to its mmol/ml value. So: 1.354920197 mmol/ml * 2 + 1.645969579 mmol/ml = 4.355809974 mmol/ml = 4.355809974 meq/ml Rounding the results to 3 significant figures gives: phosphate = 3.00 mmol/ml potassium = 4.36 meq/ml