SAT Subject Test Chemistry,10 edition

(Marvins-Underground-K-12) #1
MX 2    (s) →   M^2 +   (aq)    +   2X− (aq),

and


Ksp =   [M2+][X−]^2     =   x   (2x)^2  =   4x^3 ;  et  cetera.

In general, note how the stoichiometry comes into play twice: once to determine the relationship
between the amount of solid dissolved and the amount of ions present, and again as the exponent
in the law of mass action.


TEST STRATEGY


Predict your    answer  before  you go  to  the answer  choices so  you don’t   get persuaded   by  the
wrong choices you’ll find there. This helps protect you from persuasive or tricky wrong
answer choices, which are often logical twists on the correct choice.

Example: The solubility of Fe(OH) 3 in aqueous solution was determined to be 4.5 × 10−10 mol/L.
What is the value of the Ksp for Fe(OH) 3?
Solution: The molar solubility (the solubility of the compound in mol/L) is given as 4.5 × 10−10 M.
The equilibrium concentration of each ion can be determined from the molar
solubility and the balanced dissociation reaction of Fe(OH) 3 . The dissociation reaction
is:


Fe(OH) 3    (s) →   Fe3+    (aq)    +   3OH−    (aq)

Thus, for every mole of Fe(OH) 3 that dissociates, one mole of Fe3+ and three moles of OH− are
produced. Since the solubility is 4.5 × 10−10 M, the Ksp can be determined as follows:

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