The minimum number of grams of I that must be present is ; 0.01836 * 10⁻³ mol
<u>Given data : </u>
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
<h3>Determine the number of I that must be present </h3>
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 :<u> </u><u>determine the</u><u> molarity of I</u>
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 :<u> Determine the minimum number of grams of I </u>
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
