I don't understand the hydrolysis of Zn(NO_3)_2: why will the solution be acidic?

This is from the answers volume to my textbook: In the middle line, it's clear that we are left with two positively charged ions and two negatively charged ions - I mean 2NO3(-). Why then the solution turns acidic? Would not the positives and the negatives cancel each other out?

Feb 9, 2016

Answer:

Here's what's going on here.

Explanation:

Before going into anything else, it's important to make sure that you understand what it means to have an acidic solution.

The only indicator to go by when deciding whether or not a solution is acidic, basic, or neutral is the ratio that exists between the hydronium ions, ${\text{H"_3"O}}^{+}$, written as hydrogen ions in your book, ${\text{H}}^{+}$, and the hydroxide ions, ${\text{OH}}^{-}$.

If you have

$\left[{\text{H"_3"O"^(+)] > ["OH}}^{-}\right] \to$ the solution is acidic

$\left[{\text{H"_3"O"^(+)] < ["OH}}^{-}\right] \to$ the solution is basic

$\left[{\text{H"_3"O"^(+)] = ["OH}}^{-}\right] \to$ the solution is neutral

And that's it. The only way another anion or cation can impact the pH of a solution is if it changes the concentrations of those two ions.

Now, here's a simplified way of looking at that reaction.

Zinc nitrate, "Zn"("NO"_3)_2, is a soluble ionic compound, which means that it will dissociate completely in aqueous solution to form zinc cations, ${\text{Zn}}^{2 +}$, and nitrate anions, ${\text{NO}}_{3}^{-}$

${\text{Zn"("NO"_3)_text(2(aq]) -> "Zn"_text((aq])^(2+) + 2"NO}}_{\textrm{3 \left(a q\right]}}^{-}$

The nitrate anion is the conjugate base of a strong acid, nitric acid, ${\text{HNO}}_{3}$, which means that it will not react with water to reform the acid.

On the other hand, the zinc cation is the conjugate acid of a weak base, zinc hydroxide, "Zn"("OH")_2, which means that it will react with water to reform the weak base

$\textcolor{b l u e}{{\text{Zn"_text((aq])^(2+) + 4"H"_2"O"_text((l]) rightleftharpoons "Zn"("OH")_text(2(aq]) + 2"H"_3"O}}_{\textrm{\left(a q\right]}}^{+}}$

Notice that this reaction produces hydronium ions. This tells you that the balance that exists between hydronium cations and hydroxide anions in a neutral solution will be disrupted, since you now have more hydronium cations that hydroxide anions $\to$ the solution is acidic.

This is exactly what your book suggests, but it does so in a slightly different manner.

Here's what's going on there. You can represent zinc nitrate as

"Zn"^(2+)("NO"_3^(-))("NO"_3^(-))

and water as

"H"^(+)("OH"^(-))

When zinc nitrate is placed in aqueous solution, hydroxide anions coming from water will liberate the nitrate anions in solution. Mind you, ${\text{HNO}}_{3}$ does not form as a molecule!

In two stages, this reaction can be written as

${\text{Zn"^(2+)("NO"_3^(-))("NO"_3^(-)) + "H"^(+)("OH"^(-)) rightleftharpoons "Zn"^(2+)("NO"_3^(-))("OH"^(-)) + "H"^(+) + "NO}}_{3}^{-}$

followed by

${\text{Zn"^(2+)("NO"_3^(-))("OH"^(-)) + "H"^(+)("OH"^(-)) rightleftharpoons "Zn"^(2+)("OH"^(-))("OH"^(-)) + "H"^(+) + "NO}}_{3}^{-}$

The nitrate anions are spectator ions, which means that you can write the net ionic equation as

${\text{Zn"^(2+) + "H"^(+)("OH"^(-)) rightleftharpoons overbrace("Zn"^(2+)("OH"^(-)))^(color(purple)("ZnOH"^(+))) + "H}}^{+}$

and

${\text{Zn"^(2+)("OH"^(-)) + "H"^(+)("OH"^(-)) rightleftharpoons overbrace("Zn"^(2+)("OH"^(-))("OH"^(-)))^(color(purple)("Zn"("OH")_2)) + "H}}^{+}$

Put all this together to get

${\text{Zn"^(2+) + "H"^(+)("OH"^(-)) rightleftharpoons color(red)(cancel(color(black)("Zn"^(2+)("OH"^(-))))) + "H}}^{+}$
color(red)(cancel(color(black)("Zn"^(2+)("OH"^(-))))) rightleftharpoons "Zn"^(2+)("OH"^(-))("OH"^(-)) + "H"^(+)
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}$

$\textcolor{b l u e}{{\text{Zn"^(2+) + 2["H"^(+)("OH"^(-))] rightleftharpoons "Zn"^(2+)("OH"^(-))("OH"^(-)) + 2"H}}^{+}}$

Once again, the reaction results in the production of hydronium ions (shown here as ${\text{H}}^{+}$), which is why the resulting solution is acidic.

Feb 9, 2016

Answer:

The solution is acidic due to the following:

Explanation:

Aqueous zinc(II) ions are quite acidic as is the case with many metal ions.

In water they are surrounded by 6 water ligands in octahedral symmetery:

Charged metal ions like this have quite a high charge density. This tends to withdraw electron density from the O-H bond thus weakening it.

As a result the ion undergoes hydrolysis:

${\left[Z n {\left({H}_{2} O\right)}_{6}\right]}^{2 +} + {H}_{2} O r i g h t \le f t h a r p \infty n s {\left[Z n {\left({H}_{2} O\right)}_{5} \left(O H\right)\right]}^{+} + {H}_{3} {O}^{+}$

So you can see that acidic ${H}_{3} {O}^{+}$ are released. The nitrate ions remain as spectators.

The ion produced is signified by $Z n O {H}^{+}$ in your textbook, the 5 ${H}_{2} O$ ligands being left out.

Your textbook is very confusing for this. It shows $Z {n}_{2}^{+}$ which implies a species which consists of 2 zinc atoms bonded together with a net charge of +1.

This is clearly incorrect as charge would not be conserved on either side of the equations. It should read $Z {n}^{2 +}$ to make sense.