Answer to Question #73740 in General Chemistry for Esther

Question

How many grams of potassium phosphate are needed to be added to 40.00mL of water so that the final concentration of the potassium ions will be 0.150 M?

Answer 1

Mr(K3PO4) = 212.3 g/mol

0.15mol (K+) - 1000ml
X mol (K+) - 40ml
X = 40*0.15/1000=0.006mol (K+)
n(K3PO4) = 0.006 mol /3 = 0.002 mol
m(K3PO4) = 0.002 mol*212.3g/mol=0.425g