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

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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!

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Answer 1 · 2018-06-28T16:38:37

Is it not $B$ $B$ ?

Explanation:

The net ionic equation is simply..

$B a^{2 +} + S O_{4}^{2 -} \to B a S O_{4} (s) ⏐ ⏐ ↓$ $Ba^(2+) + SO_4^(2-) rarr BaSO_4(s)darr$ ...

The hydronium ion REMAINS in solution, i.e. $2 equiv$ $"2 equiv"$ $H_{3} O^{+}$ $H_3O^+$ ...

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...

$H_{2} O (l) ⇌ H^{+} + H O^{-}$ $H_2O(l)rightleftharpoonsH^+ + HO^-$ ...and $K_{w} = [H O^{-}] [H^{+}] = 10^{- 14}$ $K_w=[HO^-][H^+]=10^-14$ .

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

$H C l (g) H_{2} O - - \to H_{3} O^{+} + C l^{-}$ $HCl(g) stackrel(H_2O)rarrH_3O^+ + Cl^-$

A saturated solution solution of $H C l (a q)$ $HCl(aq)$ is approx $10.6 \cdot m o l \cdot L^{- 1}$ $10.6*mol*L^-1$ with respect to $H_{3} O^{+}$ $H_3O^+$ ...

And of course the $H^{+}$ $H^+$ is a label of convenience..in aqueous solution what we term $H^{+}$ $H^+$ is possibly $H_{7} O_{3}^{+}$ $H_7O_3^+$ or $H_{9} O_{4}^{+}$ $H_9O_4^+$ ...i.e. a CLUSTER of $2 - 3$ $2-3$ or more water molecules WITH AN EXTRA $H^{+}$ $H^+$ associated with the cluster. Certainly the $H^{+}$ $H^+$ can move rapidly from cluster to cluster, tunnel if you like, given the vast aqueous mobility of $H^{+}$ $H^+$ and $H O^{-}$ $HO^-$ ...

And so we represent the acidium ion as $H^{+}$ $H^+$ or even $H_{3} O^{+}$ $H_3O^+$ ...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.](https://socratic.org/questions/586bb384b72cff43cf5b0319)

Anyway, I know that $Inorganic Chemistry, Principles of$ $"Inorganic Chemistry, Principles of"$ $Structure and Reactivity, J. E. Huheey$ $"Structure and Reactivity, J. E. Huheey"$ 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...

Answer 2 · 2018-06-29T18:05:14

The correct answer is option b).

Explanation:

A substance behaves as an acid if it provides $H^{+}$ $"H"^"+"$ 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:
$H_{2} {SO}_{4} (aq) + 2NaOH(aq) \to {Na}_{2} {SO}_{4} (aq) + 2 H_{2} O(l)$ $"H"_2"SO"_4"(aq)" + "2NaOH(aq)" → "Na"_2"SO"_4"(aq)" + 2"H"_2"O(l)"$

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

Net ionic equation:
$color(red)(cancel(color(black)(2)))"H"^"+""(aq)" +color(red)(cancel(color(black)(2)))"OH"^"-""(aq)" → color(red)(cancel(color(black)(2)))"H"_2"O(l)"$

$"H"^"+""(aq)" +"OH"^"-""(aq)" → "H"_2"O(l)"$

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

b)

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

Ionic equation:
$color(red)(cancel(color(black)(2"H"^"+""(aq)"))) + "SO"_4^"2-""(aq)" + "Ba"^"2+""(aq)" + color(red)(cancel(color(black)("2Cl"^"-""(aq)"))) → "BaSO"_4"(s)" + color(red)(cancel(color(black)(2"H"^"+""(aq)"))) + color(red)(cancel(color(black)("2Cl"^"-""(aq)")))$

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

$"H"^"+"$ 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:
$"H"_2"SO"_4"(aq)" + "CuO(s)" → "CuSO"_4"(aq)" + "H"_2"O(l)"$

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

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

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

d)

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

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

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

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

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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!

2 Answers

Explanation:

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

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!

2 Answers

Explanation:

Explanation:

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Impact of this question

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!