# All the reactions are correct , so what could be the appropriate answer then?

## In which reaction is dilute sulfuric acid not behaving as an acid? a) H2SO4 + 2NaOH = Na2SO4 +2H2O b) H2SO4 + BaCl2 = BaSO4 + 2HCl c) H2SO4 + CuO = CuSO4 + H2O d) H2SO4 + Mg = MgSO4 + H2 * Pardon any tying error in the equations!

Jun 28, 2018

Is it not $B$?

#### Explanation:

The net ionic equation is simply..

$B {a}^{2 +} + S {O}_{4}^{2 -} \rightarrow B a S {O}_{4} \left(s\right) \downarrow$...

The hydronium ion REMAINS in solution, i.e. $\text{2 equiv}$ ${H}_{3} {O}^{+}$...

Just in response to a request for clarification...I will try to make what I have written less confusing... We often speak of acidity in terms of the characteristic cation of the solvent...for water we often write...

${H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}^{+} + H {O}^{-}$...and ${K}_{w} = \left[H {O}^{-}\right] \left[{H}^{+}\right] = {10}^{-} 14$.

Here, ${H}^{+}$ is CONCEIVED to be the acid principle...and $H {O}^{-}$ is the base principle, the characteristic anion... An acid dissolved in water is INCREASES the concentration of the anion, i.e. for gaseous $H C l$ we write...

$H C l \left(g\right) \stackrel{{H}_{2} O}{\rightarrow} {H}_{3} {O}^{+} + C {l}^{-}$

A saturated solution solution of $H C l \left(a q\right)$ is approx $10.6 \cdot m o l \cdot {L}^{-} 1$ with respect to ${H}_{3} {O}^{+}$...

And of course the ${H}^{+}$ is a label of convenience..in aqueous solution what we term ${H}^{+}$ is possibly ${H}_{7} {O}_{3}^{+}$ or ${H}_{9} {O}_{4}^{+}$...i.e. a CLUSTER of $2 - 3$ or more water molecules WITH AN EXTRA ${H}^{+}$ associated with the cluster. Certainly the ${H}^{+}$ can move rapidly from cluster to cluster, tunnel if you like, given the vast aqueous mobility of ${H}^{+}$ and $H {O}^{-}$...

And so we represent the acidium ion as ${H}^{+}$ or even ${H}_{3} {O}^{+}$...this is a CONCEPTION rather than the reality...but it is useful to write chemical equations invoking them as actual species... I have written much the same thing here and here.

Anyway, I know that $\text{Inorganic Chemistry, Principles of}$ $\text{Structure and Reactivity, J. E. Huheey}$ is available in most university libraries (it may even be available for download but you might have to pledge your first-born to get it), and this has a good chapter on the concept of acidity...good luck...

Jun 29, 2018

The correct answer is option b).

#### Explanation:

A substance behaves as an acid if it provides $\text{H"^"+}$ ions that get used up during a reaction.

It is easiest to see what is happening if we use net ionic equations.

a)

Molecular equation:
$\text{H"_2"SO"_4"(aq)" + "2NaOH(aq)" → "Na"_2"SO"_4"(aq)" + 2"H"_2"O(l)}$

Ionic equation:
$2 \text{H"^"+""(aq)" + color(red)(cancel(color(black)("SO"_4^"2-""(aq)"))) + color(red)(cancel(color(black)("2Na"^"+""(aq)"))) +2"OH"^"-""(aq)" → color(red)(cancel(color(black)("2Na"^"+""(aq)"))) + color(red)(cancel(color(black)("SO"_4^"2-""(aq)"))) + 2"H"_2"O(l)}$

Net ionic equation:
$\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \text{H"^"+""(aq)" +color(red)(cancel(color(black)(2)))"OH"^"-""(aq)" → color(red)(cancel(color(black)(2)))"H"_2"O(l)}$

$\text{H"^"+""(aq)" +"OH"^"-""(aq)" → "H"_2"O(l)}$

$\text{H"^"+}$ is getting used up in this reaction, so the sulfuric acid is behaving as an acid.

b)

Molecular equation:
$\text{H"_2"SO"_4"(aq)" + "BaCl"_2"(aq)" → "BaSO"_4"(s)" + 2"HCl""(aq)}$

Ionic equation:
$\textcolor{red}{\cancel{\textcolor{b l a c k}{2 \text{H"^"+""(aq)"))) + "SO"_4^"2-""(aq)" + "Ba"^"2+""(aq)" + color(red)(cancel(color(black)("2Cl"^"-""(aq)"))) → "BaSO"_4"(s)" + color(red)(cancel(color(black)(2"H"^"+""(aq)"))) + color(red)(cancel(color(black)("2Cl"^"-""(aq)}}}}$

Net ionic equation:
$\text{SO"_4^"2-""(aq)" + "Ba"^"2+""(aq)" → "BaSO"_4"(s)}$

$\text{H"^"+}$ is a spectator ion.

It is not getting used up in this reaction, so the sulfuric acid is not behaving as an acid.

c)

Molecular equation:
$\text{H"_2"SO"_4"(aq)" + "CuO(s)" → "CuSO"_4"(aq)" + "H"_2"O(l)}$

Ionic equation:
$2 \text{H"^"+""(aq)" + color(red)(cancel(color(black)("SO"_4^"2-""(aq)"))) + "CuO(s)" → "Cu"^"2+""(aq)" + color(red)(cancel(color(black)("SO"_4^"2-""(aq)"))) + "H"_2"O(l)}$

Net ionic equation:
$2 \text{H"^"+""(aq)" + "CuO(s)" → "Cu"^"2+""(aq)" + "H"_2"O(l)}$

$\text{H"^"+}$ is getting used up in this reaction, so the sulfuric acid is behaving as an acid.

d)

Molecular equation:
$\text{H"_2"SO"_4"(aq)" + "Mg(s)" → "MgSO"_4"(aq)" + "H"_2"(g)}$

Ionic equation:
$2 \text{H"^"+""(aq)" + color(red)(cancel(color(black)("SO"_4^"2-""(aq)"))) + "Mg(s)" → "Mg"^"2+""(aq)" + color(red)(cancel(color(black)("SO"_4^"2-""(aq)"))) + "H"_2"(g)}$

Net ionic equation:
$2 \text{H"^"+""(aq)" + "Mg(s)" → "Mg"^"2+""(aq)" + "H"_2"(g)}$

$\text{H"^"+}$ is getting used up in this reaction, so the sulfuric acid is behaving as an acid.