Question

0.276 g KH₂PO₄ (record the mass you use to the nearest 0.1 mg) and transfer quantitatively into a 250 mL volumetric flask. Bring to volume. Use a volumetric pipet to transfer 1.0 mL of this solution into a 100 mL volumetric flask and bring to volume. Calculate the molarity of this more dilute stock solution.

Answer 1

Molar mass of KH2PO4,

MM = 1*MM(K) + 2*MM(H) + 1*MM(P) + 4*MM(O)

= 1*39.1 + 2*1.008 + 1*30.97 + 4*16.0

= 136.086 g/mol

mass(KH2PO4)= 0.276 g

use:

number of mol of KH2PO4,

n = mass of KH2PO4/molar mass of KH2PO4

=(0.276 g)/(1.361*10^2 g/mol)

= 2.028*10^-3 mol

volume , V = 2.5*10^2 mL

= 0.25 L

use:

Molarity,

M = number of mol / volume in L

= 2.028*10^-3/0.25

= 8.113*10^-3 M

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given:

M1 = 8.113*10^-3 M

V1 = 1.0 mL

V2 = 100.0 mL

use:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (8.113*10^-3*1)/100

M2 = 8.113*10^-5 M

Answer: 8.11*10^-5 M