When a bottle of wine is opened, its hydrogen ion concentration, $[H_{3} O^{+}] = 4.1 \times 10^{- 4} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=4.1xx10^-4molL^-1$ ; what is its $p H$ $pH$ ? How does the $p H$ $pH$ evolve after the wine is left to stand?

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When a bottle of wine is opened, its hydrogen ion concentration, $[H_{3} O^{+}] = 4.1 \times 10^{- 4} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=4.1xx10^-4molL^-1$ ; what is its $p H$ $pH$ ? How does the $p H$ $pH$ evolve after the wine is left to stand?

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Answer 1 · 2017-05-07T16:44:13

$p H = - {log}_{10} [H_{3} O^{+}]$ $pH=-log_10[H_3O^+]$

Explanation:

And thus............when freshly opened $p H = - {log}_{10} (4.1 \times 10^{- 4}) = 3.39$ $pH=-log_10(4.1xx10^-4)=3.39$ .

(And in fact most wines have a $p H$ $pH$ around this level.) See here for more detail on the definition of $p H$ $pH$ .

And later $[H_{3} O^{+}] = 0.0023 \cdot m o l \cdot L^{- 1}; p H = 2.64$ $[H_3O^+]=0.0023*mol*L^-1; pH=2.64$

The wine must not have been very good, because most wine is consumed within 12 hours after opening. What likely occurred is that the ethyl alcohol air-oxidized up to acetic acid, which is a carboxylic acid, and thus likely to have a lower $p H$ $pH$ in aqueous solution. (Note that this oxidation is why we seal wine bottles with an air-tight cork/cap).

For the oxidation of ethyl alcohol to acetic acid we could write the equation:

$H_{3} {CCH}_{2} OH(aq) + O_{2} (g) \to H_{3} {CCO}_{2} H(aq) + H_{2} O(l)$ $"H"_3"CCH"_2"OH(aq)" +"O"_2"(g)" rarr "H"_3"CCO"_2"H(aq)"+"H"_2"O(l)"$

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When a bottle of wine is opened, its hydrogen ion concentration, $[H_{3} O^{+}] = 4.1 \times 10^{- 4} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=4.1xx10^-4molL^-1$ ; what is its $p H$ $pH$ ? How does the $p H$ $pH$ evolve after the wine is left to stand?

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Explanation:

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When a bottle of wine is opened, its hydrogen ion concentration, [H3O+]=4.1×10−4⋅mol⋅L−1[H_3O^+]=4.1xx10^-4*mol*L^-1; what is its pHpH? How does the pHpH evolve after the wine is left to stand?

1 Answer

Explanation:

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When a bottle of wine is opened, its hydrogen ion concentration, $[H_{3} O^{+}] = 4.1 \times 10^{- 4} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=4.1xx10^-4molL^-1$ ; what is its $p H$ $pH$ ? How does the $p H$ $pH$ evolve after the wine is left to stand?