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A solution containing lead(II) nitrate is mixed with one containing sodium bromide to form a solution that is 0.0630 M in Pb(NO3)2 and 0.0103 M in NaBr. What is the value of Q for the insoluble product? Express the reaction quotient to three significant figures.

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Answer:

Q = 6.68x10⁻⁶

Step-by-step explanation:

The initial solutions are Pb(NO₃)₂ and NaBr. As sodium and nitrates salts are soluble, the insoluble product must be the formed from the other ions, PbBr₂.

The soluble product equilibrium is written as:

PbBr₂(s) ⇄ Pb²⁺(aq) + 2Br⁻(aq)

Where Q is defined as:

Q = [Pb²⁺] [Br⁻]²

As:

[Pb²⁺] = 0.0630M

[Br⁻] = 0.0103M

Q = [Pb²⁺] [Br⁻]²

Q = [0.0630M] [0.0103M]²

Q = 6.68x10⁻⁶

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