Question

The concentration of a potassium iodide solution can be determined by first adding excess silver nitrate: I'(aq) + Ag+ (aq) → Agl(s) The excess silver ion remaining in solution is then determined by reaction with a potassium thiocyanate (KSCN) solution of known concentration: Ag (aq) + SCN(aq) → AgSCN(S) In an experiment, 50.00 ml of 0.0565 M AgNO3 was added to 25.00 ml of a potassium iodide solution. It then took 8.32 ml of 0.0510 M KSCN solution to precipitate the unreacted silver ions. What is the concentration of the original KI solution?

Answer 1

Answer:

0.096 M

Explanation:

Given:

50.00 ml of 0.0565 M AgNO₃ was added to 25.00 ml of a potassium iodide solution

8.32 ml of 0.0510 M KSCN solution to precipitate the unreacted silver ions

Now,

The moles of AgNO₃ added = Volume of solution × Concentration

or

The moles of AgNO₃ added = 0.05 × 0.0565 = 0.002825 moles

and

The number of moles of KSCN reacted

= Volume  × Concentration

= 0.00832 × 0.0510

= 0.00042432 moles

Now,

The number of moles of AgNO₃ that reacted with KI

= The moles of AgNO₃ added - The number of moles of KSCN reacted

= 0.002825 - 0.00042432

= 0.00240068

Therefore,

The concentration of the original KI solution = $\frac{\textup{Moles of KI}}{\textup{Volume of solution in liter}}$

or

The concentration of the original KI solution =  $\frac{0.00240068}{25\times10^{-3}}$

or

The concentration of the original KI solution = 0.096 M