# Chemsitry

The Ksp of PbBr2 is 6.60× 10^–6.

What is the molar solubility of PbBr2 in pure water?

What is the molar solubility of PbBr2 in 0.500 M KBr solution?

What is the molar solubility of PbBr2 in a 0.500 M Pb(NO3)2 solution?

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1. PbBr2---> Pb + 2Br

Ksp=[Pb][2Br]^2

6.60× 10^–6=[x][2x]^2

Solving for x

6.60× 10^–6=4x^3

6.60× 10^–6/4=x^3

(6.60× 10^–6/4)^1/3=x

B.)

PbBr2---> Pb + 2Br

Ksp=[Pb][2Br]^2

6.60× 10^–6=[x][0.500M]^2

Solve for x,

6.60× 10^–6/[(0.500M)^2]=x

C.)

PbBr2---> Pb + 2Br

Ksp=[Pb][2Br]^2

6.60× 10^–6=[0.500M][2x]^2

Solve for x,

6.60× 10^–6/[(0.500M)^2]=x

6.60× 10^–6/(0.500M)=4x^2

sqrt*{[6.60× 10^–6/(0.500M)]/4}=x

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2. I don't disagree with any of the answers. Technically, and I realize this is nit-picking, I disagree with the set up. As an example,
PbBr2 ==> Pb^2+ + 2Br^-
But Ksp says "the product of the molar solubility of the ions, each raised to the power indicated in the balanced equation, is a constant". I would write that as Ksp = (Pb^2+)(Br^-)^2. It isn't twice the bromide ion x lead ion, it is THE bromide x lead ion. Yes, Pb ion is x and Br is 2x but 2x is THE bromide ion, not twice the bromide, but it is twice the PbBr2 that dissolves which is why it's 2x to begin with.

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3. I understand. That is why I worked on a problem and developed a headache when I couldn't figure out why the answer doubled compared to the answer that I calculated. If I set up the way that you mentioned then I would not have ran into that problem, and I can prevent others from performing the same mistake. Thank you for the advice. Also, I do not consider it nit-picking. I always thought it was nick-picking. We learn something new everyday.

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4. I lent my sleeping bag to a co-worker once, when I used it the next night I woke up suddenly, feeling something crawling in my crotch. I spent the next 2 hours picking the critters off. then I went cross eyed picking nits. I got them all. Unpicked nits can grow into a HUGE problem. Never think you are being to thorough, especially in chemistry!!

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5. t,mgbfjp[o;['

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