Question

If 30 g of potassium phosphate are diluted with enough water to make 1.5 dm³ of a stock solution, how much of the stock solution would be needed to make 100 mL of a 0.015 M solution of potassium ions?

Answer is 5.3 mL, but please explain.

Answer 1

1st calculate the initial concentration

Molar mass of K3PO4 = 3*MM(K) + 1*MM(P) + 4*MM(O)

= 3*39.1 + 1*30.97 + 4*16.0

= 212.27 g/mol

mass of K3PO4 = 30 g

we have below equation to be used:

number of mol of K3PO4,

n = mass of K3PO4/molar mass of K3PO4

=(30.0 g)/(212.27 g/mol)

= 0.1413 mol

volume , V =1.5 dm^3 = 1.5 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 0.1413/1.5

= 9.422*10^-2 M

use dilution formula

M1*V1 = M2*V2

Here:

M1 is molarity of solution before dilution

M2 is molarity of solution after dilution

V1 is volume of solution before dilution

V2 is volume of solution after dilution

we have:

M1 = 0.09422 M

M2 = [K3PO4] = [K+]/3 = 0.015/3 = 0.005 M

V2 = 100.0 mL

we have below equation to be used:

M1*V1 = M2*V2

V1 = (M2 * V2) / M1

V1 = (0.005*100)/0.0942

V1 = 5.31 mL

Answer: 5.31 mL