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Olegator [25]
3 years ago
12

The concentration of pb2+ in a solution saturated with pbbr2(s) is 2.14 ✕ 10-2 m. calculate ksp for pbbr2.

Chemistry
2 answers:
Sergio [31]3 years ago
8 0

Answer : The value of K_{sp} is, 3.92\times 10^{-5}

Explanation :

The balanced equilibrium reaction will be,

PbBr_2\rightleftharpoons Pb^{2+}+2Br^{1-}

The expression for solubility constant for this reaction will be,

K_{sp}=(s)\times (2s)^2

K_{sp}=4(s)^3

Now put the value of solubility in this expression, we get the value of K_{sp}

K_{sp}=4(2.14\times 10^{-2})^3

K_{sp}=3.92\times 10^{-5}

Therefore, the value of K_{sp} is, 3.92\times 10^{-5}

Ugo [173]3 years ago
3 0
Concentration = 2.14 âś• 10-2 m 
For [Br-], there are 2 ions so 2 x 2.14 x 10^-2 =4.28 x 10^-2  
Ksp = [Pb][Br]^2 = 2.14 âś• 10-2 x (4.28 x 10^-2 )^2 = 39.20 x 10^-6 
Ksp = 3.92 x 10^-5
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