A 50.0 mL sample of KCl requires 22.40 mL of 0.0229 M Pb(NO3)2 in order to completely titrate it. What is the Molarity of the KCl sample?

0.02052M

Explanation:

First, we need to write a balanced equation for the reaction. This is illustrated below:

2KCl + Pb(NO3)2 → 2KNO3 + PbCl2

The following were obtained from the question:

Molarity of Pb(NO3)2 = 0.0229M

Volume of Pb(NO3)2 = 22.40 = 22.4/1000 = 0.0224L

Number of mole of Pb(NO3)2 =?

Recall:

Mole = Molarity x Volume

Mole of Pb(NO3)2 = 0.0229x0.0224

Mole of Pb(NO3)2 = 5.13x10^-4mole

From the equation,

1mole of Pb(NO3)2 required 2moles KCl.

Therefore, 5.13x10^-4mole of Pb(NO3)2 will require = 5.13x10^-4x2 = 1.026x10^-3mole of KCl.

Now we can use this amount (i.e 1.026x10^-3mole) to find the molarity of KCl. This is illustrated below:

Mole of KCl = 1.026x10^-3mole

Volume of KCl = 500mL = 50/1000 = 0.05L

Molarity =?

Molarity = mole /Volume

Molarity of KCl = 1.026x10^-3/0.05

Molarity of KCl = 0.02052M

it should be p1 v1 = p2 v2