Question #7e83a

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Question #7e83a

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Answer 1 · 2017-02-07T11:58:38

The concentration of $H^{+}$ $H^+$ ion will be the same as the stated concentration of the solution that contains the strong acid.

Explanation:

If an acid is said to be strong, this means that 100% of the molecules of the acid undergo ionization. For example, in the case of HCl

$H C l \to H^{+} + C l^{-}$ $HCl rarr H^+ + Cl^-$

This ionization of the HCl molecules is complete. There are no HCl molecules remaining in the solution after this has happened (which takes place in a very short time period).

So, if we place 0.10 mole of HCl into 100 mL of water, we will find the solution contains 0.10 mole of $H^{+}$ $H^+$ ions. The concentration of the ions is found in the usual way:

$[H^{+}] = \frac{0.10 mol}{0.100 L} = 1.0 \frac{mol}{L}$ $[H^+] = (0.10 "mol")/(0.100 L) = 1.0 "mol"/L$

Likewise, if the label on the bottle reads "0.200 M HCl" (or any other strong acid for that matter), the $[H^{+}]$ $[H^+]$ will be 0.200 M.

Before trying to calculate the $[H^{+}]$ $[H^+]$ in a solution, it is essential that you know whether the acid is strong or weak, as the calculation is done in very different ways for these two cases.

By the way, if the acid is "diprotic", meaning there are two $H^{+}$ $H^+$ ions that ionize, only the first ionization is strong; the second is weak.

For example, $H_{2} S O_{4}$ $H_2SO_4$ :

$H_{2} S O_{4} \to H^{+} + H S O_{4}^{-}$ $H_2SO_4 rarr H^+ + HSO_4^-$ is strong

$H S O_{4}^{- \to} H^{+} + S O_{4}^{2 -}$ $HSO_4^- rarr H^+ + SO_4^(2-)$ is weak

So, a 0.100 M solution of $H_{2} S O_{4}$ $H_2SO_4$ has a #[H^+] that is a bit higher than 0.100 M, but not 0.200 M. (This is a more difficult calculation than the one you are looking at.)

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Question #7e83a

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