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The solubility of an ionic compound, MX (molar mass of 346g) is #4.63*10^-3 g/L#. What is Ksp for the compound?

Answer 1

Cameron Andrews

#K_(sp)~=1.8xx10^-10#

#MX(s) rightleftharpoonsM^(+) + X^(-)#

Since this reaction is in equilibrium, we write the equilibrium expression as usual:

#K_(sp) = [M^+][X^-]#

#[MX]# does not appear in the expression inasmuch as we cannot speak of the concentration of a solid.

Now #S# #=# #4.63xx10^(-3)*g*L^-1#.

#=# #(4.63xx10^(-3)*g)/(346*g*mol*L^-1)# #=# #1.34xx10^-5*mol*L^-1# with respect to #MX#.

Since #[MX]# #=# #[M^+]# #=# #[X^-]#,

#K_(sp)=[M^+][X^-]# #=# #(1.34xx10^-5)^2#

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Answer 2

Mateo Aguilar

[ K_{sp} = \dfrac{[\text{M}^{n+}][\text{X}^{m-}]}{s} ]

[ K_{sp} = \dfrac{(4.63 \times 10^{-3})^2}{346} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.