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Ne4ueva [31]
3 years ago
7

At 25 °C, only 0.0990 mol of the generic salt AB3 is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C? AB3(s)↽−−

⇀A3+(aq)+3B−(aq)
Chemistry
1 answer:
maksim [4K]3 years ago
4 0

Answer:

Ksp=2.59x10^{-3}

Explanation:

Hello,

In this case, given the 0.0990 moles of the salt are soluble in 1.00 L of water only, we can infer that the molar solubility is 0.099 M. Next, since the dissociation of the salt is:

AB_3\rightleftharpoons A^{3+}+3B^-

The concentrations of the A and B ions in the solution are:

[A]=0.099 \frac{molAB_3}{L}*\frac{1molA}{1molAB_3}  =0.0099M

[B]=0.099 \frac{molAB_3}{L}*\frac{3molB}{1molAB_3}  =0.000.297M

Then, as the solubility product is defined as:

Ksp=[A][B]^3

Due to the given dissociation, it turns out:

Ksp=[0.099M][0.297M]^3\\\\Ksp=2.59x10^{-3}

Regards.

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