Question

In a similar experiment, the conductimetric titration of 15.00 mL of an unknown lead(II) nitrate solution with a 0.105 M potassium iodide solution was carried out. The volume of potassium iodide solution required to reach the equivalence point was found to be 17.35 mL. Calculate the molarity of the unknown lead(II) nitrate solution.

Answer 1

0.060725M

Step-by-step explanation:

We first write the reaction equation which is as follows;

Pb(NO3)2(aq) + 2KI(aq) ==> PbI2(s) + 2KNO3(aq)

From here, we can see 1 mole of the nitrate reacted with 2 moles of the iodide.

We can use the standardization formula to get this.

CaVa/Na = CbVb/Nb

Given:

Ca =? Va= 15.00ml Cb = 0.105M Vb = 17.35ml Na = 1 , Nb = 2

We substitute these values into the equation :

(Ca × 15)/1 = ( 0.105 × 17.35)/2

15Ca = 0.910875

Ca = 0.910875/15 = 0.060725M