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
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
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