Question

I don't understand the hydrolysis of $Zn{(N{O_3})_2}$ : why will the solution be acidic?
This is from the answers volume to my textbook:
$Zn{(N{O_3})_2} + HOH \rightleftharpoons Zn(OH)N{O_3} + HN{O_3}$
$Z{n_2}^ + + 2NO_3^ - + HOH \rightleftharpoons ZnO{H^ + } + 2NO_3^ - + {H^ + }$
$Zn_2^ + + HOH \rightleftharpoons ZnO{H^ + } + {H^ + }$
In the middle line, it's clear that we are left with two positively charged ions and two negatively charged ions - I mean $2NO_3^ - $ . Why then does the solution turn acidic? Would not the positives and the negatives cancel each other out?

Answer 1

Hint: To know whether a solution is acidic, basic or neutral, we’ll have to know the $[{H_3}{O^ + }]\& [O{H^ - }]$ . The acidity and basicity can be solely decided as:
$[{H_3}{O^ + }] > [O{H^ - }] \to $the solution will be acidic
$[{H_3}{O^ + }] < [O{H^ - }] \to $ the solution will be basic
$[{H_3}{O^ + }] = [O{H^ - }] \to $ the solution will be neutral.

Complete answer:
The given compound zinc nitrate is a water-soluble ionic compound which means that it will dissociate completely in aqueous solution to give zinc cations and nitrate anions. The dissociation can be shown as:
$Zn{(N{O_3})_2} \rightleftharpoons Z{n_{ + 2}} + 2NO_3^ - $
The nitrate ion is the conjugate base of a string acid $(HN{O_3})$ and the zinc cation is the conjugate acid of a weak base $(Zn{(OH)_2})$ . Note that the strong acid will not react with water to reform the acid. But the weak base will react with water to reform the weak base $(Zn{(OH)_2})$. The reaction can be given as:
$Z{n^{ + 2}} + 4{H_2}O \rightleftharpoons Zn{(OH)_2} + 2{H_3}{O^ + }$
As you can see in this reaction, hydronium ions are formed. The concentration of the Hydronium ions increases in the water and the neutral solution disrupts. The solution will become Acidic due to increased concentration of ${H_3}{O^ + }$
Alternatively, this can be shown as follows. Zinc nitrate and water can be represented as:
$Z{n^{ + 2}}(NO_3^ - )(NO_3^ - )$ and ${H^ + }(O{H^ - })$ respectively. When zinc nitrate is dissolved in water to form an aqueous solution, the hydroxide ions of water will liberate nitrate ions in the solution. As we have said earlier, strong acid $(HN{O_3})$ will not reform. There are two stages of this reaction:
$Z{n^{ + 2}}(NO_3^ - )(NO_3^ - ) + {H^ + }(O{H^ - }) \rightleftharpoons Z{n^{ + 2}}(O{H^ - })(NO_3^ - ) + {H^ + } + NO_3^ - $
$Z{n^{ + 2}}(NO_3^ - )(O{H^ - }) + {H^ + }(O{H^ - }) \rightleftharpoons Z{n^{ + 2}}(O{H^ - })(O{H^ - }) + {H^ + } + NO_3^ - $
The nitrate ions do not react anywhere in the reaction and thus are ‘spectator ions’. Without the spectator ions the reaction can be written as:
$Z{n^{ + 2}} + {H^ + }(O{H^ - }) \rightleftharpoons Z{n^{ + 2}}(O{H^ - }) + {H^ + }$
$Z{n^{ + 2}}(O{H^ - }) + {H^ + }(O{H^ - }) \rightleftharpoons Z{n^{ + 2}}(O{H^ - })(O{H^ - }) + {H^ + }$
The overall final equation thus can be written as: $Z{n^{ + 2}} + 2[{H^ + }(O{H^ - })] \rightleftharpoons Z{n^{ + 2}}(O{H^ - })(O{H^ - }) + 2{H^ + }$
This resulting reaction also shows the formation of Hydronium ions, which makes the solution acidic.

Note:
This can also be explained based on coordination chemistry. Zn when dissolved in water form a complex with 6 water molecules. The complex formed will be $[Zn{({H_2}O)_6}]$ . On hydrolysis one water ligand is replaced with hydroxide ligand and a hydronium molecule is released, making the solution acidic.
$Zn[{({H_2}O)_6}] + {H_2}O \rightleftharpoons {[Zn{({H_2}O)_5}(OH)]^ + } + {H_3}{O^ + }$