Question

If a solution contains 1% KH2PO4 (MW 136.1) and 1% K2HPO4 (MW 174.2) what is the concentration of potassium?

Please explain how to solve it and the answer must be:

About 0.19 N

About 0.19 mEq per ml

Answer 1

per 100 g we have 1g KH2PO4

Moles of KH2Po4 = mass /molar mass of it = ( 1g) / ( 136.1g/mol) = 0.00735 mol

Moles of K2HPO4 = ( 1g/ 174.2g/mol) = 0.00574 mol ,

Moles of K = 2 x 0.00574= 0.01148

Total moles of K = 0.01148+0.00735 = 0.01883 = total equivalents of K

( since for +1 charged species moles = equivalents)

solution mass =100 g , ( assuming density = 1g/ml )

volume = mass / density = 100 g / 1g/l = 100 ml = 0.1L

Normality of K = Number of equivalents of K / solution volume in L = 0.01883 /0.1 = 0.1883 = 0.19 N

equiavlents = 0.019 per 100 ml

per 1ml = 0.019 / 100 = 0.00019 eq/ml = 0.19 meq/ml