Question

You mix 265.0 mL of 1.20 M lead (II) Nitrate with 300.0 mL of 1.55 M Potassium Iodide. The lead (II) Iodide is insoluble. Calculate (a)the final concentration of Pb²⁺ . (b) Mass of lead (II) Iodide (c) The final concentration of K⁺ (d) the final concentration of NO₃^-.

Answer 1

The equation for the reaction is;

$Pb(NO_3)_{2{(aq)}}+2KI_{(aq)}\to PbI_{2{(s)}}+2KNO_{3{(aq)}}$

a) $[Pb^{2+}]=$ $0.30.L\times 1.55M\over0.265.L\times 1.20M$ $=1.462mol.L$

b) $Mass=n\times M$

$n=1.86moles$

$M=461g/mol$

$Mass(g)=1.86Moles\times 461g/mol=870.48g$

c) $[K^+]=$ $(0.35.L\times1.86mol)\times 2 =1.302mol/L$

d) $[NO^-_3]=$ $0.265.L\times 1.20M\over0.30.L\times1.55M$ $=0.683mol.L$

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