1. The solute $\ce{AgBr}$ is least soluble in

    • A $\ce{H2O}$
    • B $1.0~\mathrm{M}\ \ce{FeBr3}$
    • C $1.0~\mathrm{M}\ \ce{CaBr2}$
    • D $1.0~\mathrm{M}\ \ce{AgNO3}$

I'm confused as to why the answer is B. Is it because $\ce{Ag}$ has low solubility with $\ce{Br}$, and the higher the concentration of $\ce{Br-}$ ions the less $\ce{Ag}$ will dissolve?


While solubilities are typically quoted in $\mathrm{g/l}$ or related units, the actual physicochemical idea behind it is an equilibrium constant known as the solubility product. In general, dissolution and precipitation can be considered two directions of the following equilibrium:

$$\ce{m A^n+ (aq) + n B^m- (aq) <<=> A_mB_n (aq) <=>> A_mB_n (s) v}\tag{1}$$

The outer sides are both preferred with respect to the central one; thus, the central compound is usually ignored, giving the following equilibrium constant:

$$K_\mathrm{sp}^{-1} = \frac{a(\ce{A_mB_n})}{a(\ce{A^n+})^m a(\ce{B^m-})^n}\tag{2}$$

Notes concerning equation $(2)$:

  • technically, this would be a precipitation equation; the solubility product is derived from the opposite equation hence the superscript $-1$.

  • activities should be used here, and not concentrations. This makes a difference as it allows for an activity of a solid (which has no concentration in solution — being solid outside of the solution).

Going the dissolution pathway, the activities of $\ce{A^n+}$ and $\ce{B^m-}$ are in the numerator and the solid’s activity is in the denominator. Thankfully, the activity of any solid compound is always $1$ per definition. That gives us $(2')$:

$$K_\mathrm{sp} = a(\ce{A^n+})^ma(\ce{B^m-})^n\tag{$2'$}$$

Which is the way the solubility product is typically represented. Now since this is an equilibrium process, Le Chatelier’s principle influcences it. Most importantly, you can add any one reactant to generate more of the product and vice-versa. If you take a look back at $(1)$, you’ll notice that the dissoluted ions enter the equation separately. Thus, we can use higher concentrations of one ion to reduce the overall solubility since then less of the other ion can remain dissolved.

This already tells us that all of options B through D result in a lower solubility since bromide ions (or silver ions) are dissolved. But how do we compare the three? The key here is that a $\pu{1M}$ solution of a salt does not necessarily mean that all its ions have a concentration of $\pu{1M}$. Instead, for $\ce{CaBr2}$ two moles of bromide are in solution for each mole of salt so if $c(\ce{CaBr2}) = \pu{1M}$, the concentration of bromide will be $c(\ce{Br-}) = \pu{2M}$.

Therefore, the solubility of $\ce{AgBr}$ will be lower in $\ce{CaBr2}$ (twice the concentration of bromide) and lower still in $\ce{FeBr3}$ (thrice the concentration of bromide). Hence B is your answer.

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  • $\begingroup$ I would claim the opposite, I think. Solubility is fundamentally how much substance $\ce{A}$ dissolves in $\ce{B}$ for specified conditions, and one way (not unique) of expressing analytical composition has dimensions $\mathsf{M\ L^{-3}}$. The solubility product is a method of estimating solubilities, but a number of criteria must be satisfied in order for $K_\mathrm{sp}$ to be useful. // 10.1021/ed018p126; solubility in Gold Book $\endgroup$ – Linear Christmas Feb 21 '18 at 12:15

AgBr would not dissolve anymore if the solubity product is greater than Ksp.In option B, that thing is equivalent to saying 3M Br-. Let AgBr dissolve by x moles/litre.
In option D , you have 1M Ag+, so if y moles/lit dissolve then
Clearly x is less than y .so answer is B

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