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Match the following:
$\begin{array}{|c|c|c|c|} \hline &\textrm{Column-I}&&\textrm{Column-II} \\\hline \textrm{(A)}&\ce{AgBr}&\mathrm{(p)}&\mathrm{(Solubility~in~water~is~more~than~expectation)} \\ \hline \textrm{(B)}&\ce{AgCN}&\mathrm{(q)}&\mathrm{(Solubility~in~acidic~solution~is more~than~in~pure~water~\\(consider~no~common~ion~effect~from~anion~of~acid)}\\ \hline \textrm{(C)}&\ce{Fe(OH)3}&\mathrm{(r)}&\mathrm{(Solubility~in~strongly~basic~solution~is~more~than~in ~pure~water)}\\ \hline \textrm{(D)}&\ce{Zn(OH)2}&\mathrm{(s)}&\mathrm{(Solubility~decreases~in~presence~of~common~anion)}\\ \hline \end{array}$

Link to original image of the question.

The problem is to 'match' options in the two columns (possibly having multiple answers?).

How to determine the solubility?

I tried this through reaction like

$\ce{AgBr + H2O -> AgOH + HBr}$

How do I know its solubility is greater or smaller than expectation?

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  • $\begingroup$ Feel free to add your attempt (a thought process), and what have you tried/read/analyzed to solve the problem. $\endgroup$
    – andselisk
    Mar 31, 2020 at 11:29

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A is wrong. $\ce{AgBr}$ is one of the least soluble compounds of $\ce{Ag}$. Why don't you have a look in the table of solubility products to be convinced? Further more, $\ce{AgBr}$ will never react with $\ce{H_2O}$ to produce a strong acid like $\ce{HBr}$ and a base like $\ce{AgOH}$. The reaction goes the other way round. Strong acids react with hydroxydes to produce a salt and water.

C is wrong, because of the Le Chatelier's principle. Try to get the reason your self.

To solve D, you must know that $\ce{Zn(OH)_2}$ is soluble in an excess of $\ce{OH^-}$ ions

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    $\begingroup$ The point is to match up entries in the two columns. $\endgroup$
    – Buck Thorn
    Aug 21, 2021 at 7:28
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    $\begingroup$ @Buck Thorn. OK. I know but I thought we should not give the whole answers without the author showing us some beginning of reasoning. So let me give him two partial answers. AgBr is poorly soluble. But its solubility decreases in if $\ce{Br-}$ ions are in excess, due to its solubility product, and the absence of any complexation reactions. So $A$ is related to $s$. Now the only product which is soluble in alcaline medium is $\ce{Zn(OH)2}$, giving the ion$\ce{[Zn(OH)4]^{2-}}$. So $D$ is related to $r$. The next substances are easy to match. $\endgroup$
    – Maurice
    Aug 22, 2021 at 10:15

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