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.
What is the minimum number of grams of I− that must be present in order for PbI2(s) ( Ksp=8.49×10−9 ) to form?

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batolisis batolisis · Answer 1 · 2022-04-04T14:04:02+02:00

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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