Suppose 30.0 mL of 0.100 M Pb(NO3)2 is added to 60.0 mL of 0.150 M KI.

Question

Suppose 30.0 mL of 0.100 M Pb(NO₃)₂ is added to 60.0 mL of 0.150 M KI. What will be the molar concentrations of all the ions in the mixture after equilibrium has been reached?

Answer

2 KI(aq) + Pb(NO₃)₂(aq) → 2 KNO₃(aq) + PbI₂(s)

This is a double replacement reaction (K and Pb swap positions).

KNO₃ is (aq) because all Nitrates are soluble (see solubility rules).

PbI₂ is (s) (see solubility rules).

Moles of KI = Volume x Molarity = 0.06 L x 0.150 = 0.009 moles

Moles of Pb(NO₃)₂ = Volume x Molarity = 0.03 L x 0.100 = 0.003 moles

Pb(NO₃)₂ is the limiting reactant.

All Pb ions will be in the form PbI₂(s) → [Pb²⁺] = 0.000 M

Moles of I- left = 0.009 moles – moles used to make PbI₂

[I^–] = Moles of I- left /Final volume

[I^–] = (0.009 – 2 x 0.003) /(0.09) = 0.033 M

[I^–] = 0.033 M

K⁺ stays soluble:

[K⁺] = 0.009/0.09 = 0.100 M

[K⁺] = 0.100 M

NO₃^– stays soluble:

[NO₃^–] = 2 x 0.003 /0.09 = 0.067 M

[NO₃^–] = 0.067 M

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