Question

Suppose that a 10-mL sample of a solution is to be tested for I− ion by addition of 1 drop (0.2 mL) of 0.13 M Pb(NO3)2.

Answer 1

The minimum number of grams of I that must be present is ; 0.01836 * 10⁻³ mol

Given data :

Volume of solution to be tested for I-ion = 10 mL

Volume of  Pb(NO₃)₂ = 0.2 mL

molarity of Pb(NO₃)₂ = 0.13 M

Determine the number of I that must be present

First step : calculate conc of PB²⁺ ions in the solution

conc of PB²⁺ ions =  ( molarity of Pb(NO₃)₂ * volume of  Pb(NO₃)₂ ) / ( total volume )

                              = ( 0.13 * 0.2 ) / ( 10 + 0.2 )

                              = ( 0.026 ) / ( 10.2 )  = 0.002549 M

Next step : determine the molarity of  I

using the dissociation reaction of PbI₂

PbI₂(s) ---> Pb²⁺ (aq)  + 2I (aq)

also;  Ksp = [ Pb²⁺ ] [ I ]²  ---- ( 1 )

From the question the given value of Ksp = 8.49 * 10⁻⁹

Therefore equation ( 1 ) becomes

8.49 * 10⁻⁹ = ( 0.002549 ) * [ I ]²

[ I ] = √ ( 8.49 * 10⁻⁹ ) / ( 0.002549 )

      = 0.0018 M

Final step : Determine the minimum number of grams of I

moles of I = molarity of I * total volume

                 = 0.0018 M * 10.2 mL

                 = 0.01836 * 10⁻³ mol

Hence we can conclude that The minimum number of grams of I that must be present is ; 0.01836 * 10⁻³ mol

Learn more about Pb(NO₃)₂ : brainly.com/question/25071409