Answer to Question #123545 in Organic Chemistry for Chan

Question

Calculate pI for the titration of 50.0 mL of 0.0250 M KI after adding 25.0 ml of
0.0500 M AgNO3.

Answer 1

Moles of KI =(50/1000) × 0.025 mole.

Moles of AgNO₃ = ( 25/1000) × 0.05 mole .

KI + AgNO₃ ------ AgI + KNO₃ .

After Reaction of KI and AgNO₃ Product formed are AgI and KNO₃ moles of AgI and KNO₃ formed are 1.25 × 10 ^-3 moles. And both KI and AgNO₃ are totally consumed .

Concentration of AgI = (1.25 / 75 ) mole / litre .

So , Concentration of I = (1.25 / 75 ) M.

So , PI = -log l = -log 1.25/75 = 1.78 . Ans...