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In a book, I read two quite contradictory statements.:

  1. $\ce{HIO > HBrO > HClO}$ wrt acidity while

  2. $\ce{HF > HCl > HBr}$.

Aren't they contradictory? If not can anyone please explain the accurate reasoning?

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    $\begingroup$ HF is a weaker acid than HCl and HBr. The sign of inequality must be reverse in the second statement. $\endgroup$ – Tejas Mar 3 '15 at 2:31
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$$\ce{HOCl>HOBr>HOI}$$ $\mathrm{p}K_a(\ce{HOCl})=7.5$, $\mathrm{p}K_a(\ce{HOBr})=8.6$, $\mathrm{p}K_a(\ce{HOI})=10.6$

The $\ce{H-O}$ bond in those oxo-acides ionizes more readily when the oxygen atom is bonded to a more electronegative atom.

  • The conjugate base $\ce{OCl-}$ is more stable than $\ce{OBr-}$. Because $\ce{Cl}$ being more electronegative, is better able to accommodate the negative charge.
  • The conjugate base $\ce{OBr-}$ is more stable than $\ce{OI-}$. Because $\ce{Br}$ being more electronegative, is better able to accommodate the negative charge.

$$\ce{HI>HBr>HCl\gg HF}$$ $\mathrm{p}K_a(\ce{HF})=3.1$, $\mathrm{p}K_a(\ce{HCl})=-6.0$, $\mathrm{p}K_a(\ce{HBr})=-9.0$, $\mathrm{p}K_a(\ce{HI})=-9.5$

$\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ are all strong acids, whereas $\ce{HF}$ is a weak acid.

As halogen size increases (as we move down the periodic table), $\ce{H-X}$ bond strength decreases and acid strength increases (a weaker bond produces a stronger acid, and vice versa).

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  • $\begingroup$ While in mhchem mode the subscript has to be encased in curly brackets, otherwise it does not work. The proper typesetting is also \mathrm{p}K_a(\ce{...}). $\endgroup$ – Martin - マーチン Mar 12 '15 at 9:29
  • $\begingroup$ As an add-on to this excellent answer: acidity and basicity are experimentally determined. The order of acidity and basicity is based on countless experiments that establish. We are left as scientists to explain why the order is such as it is. The how is the easier part - experiment! $\endgroup$ – Ben Norris Mar 12 '15 at 10:39
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I'm not sure about the first order, but I'll try to explain the reason for the second one. The acidic strength increases as $$\ce{HF<HCl<HBr}$$

One reason is the bond strength between the hydrogen and the halogen atom. Since the bond between $\ce F$ and $\ce H$ is very strong, the molecule does not easily lose $\ce H^+$. On the other hand, as the size of the halogen atom increases, the overlapping of the orbitals between halogen and hydrogen becomes less efficient. Thus, $\ce {Cl}$ and $\ce {Br}$ can more easily lose $\ce {H^+}$ than $\ce {F}$. Hence, the acidic strength increases as size of halogen atom increases.

Second reason is the stability of formation of conjugate base. When an acid loses $\ce {H^+}$ ion, it forms what is called as conjugate base. Similarly, when a base loses $\ce {OH^-}$ ion, it forms a conjugate acid. In case of $\ce {HF}$, we have $$\ce{HF <=> H^+ +F- }$$ Here, $\ce {F^-}$ is the conjugate base. Now, due to small size of $\ce {F}$, it is difficult for $\ce {F-}$ to hold negative charge. There are strong repulsions.

Now, for $\ce {HBr}$, we have $$\ce{HBr <=> H^+ +Br^-}$$

$\ce {Br}$ has a greater atomic radius. It is also less electronegative than $\ce {F}$. Hence, the repulsions are less in $\ce {Br-}$. Thus, $\ce {Br^-}$ is quite stable. The more stable is the conjugate base, the more readily the acid loses $\ce {H^+}$ ion, and so, more is its acidic strength.

Check this link. Maybe you'll find it useful.

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  • $\begingroup$ There should be an equilibrium sign in that reaction. I don't know how to put it. :p $\endgroup$ – Tejas Mar 6 '15 at 3:26
  • $\begingroup$ Use <=> for equilibrium arrows $\ce{<=>}$. If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details. $\endgroup$ – Martin - マーチン Mar 12 '15 at 9:37
  • $\begingroup$ en.wikipedia.org/wiki/Hydrogen_fluoride (also @Karan Singh) $\endgroup$ – Mithoron Mar 12 '15 at 12:49
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As Tejas said the sign of inequality in the second statement should be opposite. This is because in HF there is extensive hydrogen bonding which tends to reduce its acidic character. The reason being the high electro negativity of Fluorine. The extent of hydrogen bonding would decrease down the group due to decrease in electro negativity. A similar explanation may also be extended to reason the first one. Your book is definitely incorrect.

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  • $\begingroup$ Hey Thanks! But i am still not getting it precisely. Can you pls explain it more elaborately :) ? $\endgroup$ – girl.kriti Mar 5 '15 at 7:07

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