How would you write the chemical equation for how the acetate buffer neutralizes excess acid, $H_{3} O^{+}$ $H_3O^+$ ? How do you write the chemical equation for how the acetate buffer neutralizes excess base, $O H^{-}$ $OH^-$ ?

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How would you write the chemical equation for how the acetate buffer neutralizes excess acid, $H_{3} O^{+}$ $H_3O^+$ ? How do you write the chemical equation for how the acetate buffer neutralizes excess base, $O H^{-}$ $OH^-$ ?

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Answer 1 · 2017-07-06T18:58:21

Well, it is a buffer, the which moderates GROSS changes in $p H$ $pH$ .

Explanation:

The parent acid undergoes the equilibrium........

$H O A c (a q) + H_{2} O (l) ⇌ H_{3} O^{+} +^{-} O A c$ $HOAc(aq)+H_2O(l)rightleftharpoonsH_3O^+ + ""^(-)OAc$ ,

i.e. ${HOAc = HO(O=)CCH}_{3}$ $"HOAc = HO(O=)CCH"_3$

And thus if there are significant concentrations of acetic acid, AND acetate ions.......then if there is excess acid......

$H_{3} O^{+} +^{-} O A c \to H O A c (a q) + H_{2} O (l)$ $H_3O^+ + ""^(-)OAc rarrHOAc(aq) + H_2O(l)$

..........and if there is excess base......

$H O^{-} + H O A c (a q) \to H_{2} O (l) +^{-} O A c$ $HO^(-) + HOAc(aq) rarr H_2O(l) + ""^(-)OAc$

As to the background. The weak acid $H O A c$ $HOAc$ undergoes an acid base equilibrium in water according to the equation:

$H O A c (a q) + H_{2} O (l) ⇌ H_{3} O^{+} +^{-} O A c$ $HOAc(aq) + H_2O(l) rightleftharpoons H_3O^+ + ""^(-)OAc$

As with any equilibrium, we can write the equilibrium expression:

$K_{a}$ $K_a$ $=$ $=$ $\frac{[H_{3} O^{+}] [^{-} O A c]}{[H O A c]}$ $([H_3O^+][""^(-)OAc])/([HOAc])$

This is a mathematical expression, which we can divide, multiply, or otherwise manipulate PROVIDED that we do it to both sides of the expression. Something we can do is to take ${log}_{10}$ $log_10$ of BOTH sides.

${log}_{10} K_{a} = {log}_{10} [H_{3} O^{+}] + {log}_{10} {\frac{[^{-} O A c]}{[H O A c]}}$ $log_10K_a=log_10[H_3O^+] + log_10{([""^(-)OAc]]/[[HOAc]]}$

(Why? Because ${log}_{10} A B = {log}_{10} A + {log}_{10} B$ $log_10AB=log_10A +log_10B$ .)

Rearranging,

$- {log}_{10} [H_{3} O^{+}] - {log}_{10} {\frac{[^{-} O A c]}{[H O A c]}} = - {log}_{10} K_{a}$ $-log_10[H_3O^+] - log_10{[[""^(-)OAc]]/[[HOAc]]}=-log_10K_a$

But BY DEFINITION, $- {log}_{10} [H_{3} O^{+}] = p H$ $-log_10[H_3O^+]=pH$
, and $- {log}_{10} K_{a} = p K_{a}$ $-log_10K_a=pK_a$ .

Thus $p H = p K_{a} + {log}_{10} {\frac{[^{-} O A c]}{[H O A c]}}$ $pH=pK_a+log_10{[[""^(-)OAc]]/[[HOAc]]}$

For $acetic acid$ $"acetic acid"$ , $p K_{a} = 4.76$ $pK_a=4.76$ . And thus if there are significant quantities of acetic acid and acetate anion, the $p H$ $pH$ of the solution should be tolerably close to $4.76$ $4.76$ , i.e. close to $p K_{a}$ $pK_a$ . If there are EQUAL concentrations of acetic acid and acetate, $p H = p K_{a}$ $pH=pK_a$ . Why?

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How would you write the chemical equation for how the acetate buffer neutralizes excess acid, $H_{3} O^{+}$ $H_3O^+$ ? How do you write the chemical equation for how the acetate buffer neutralizes excess base, $O H^{-}$ $OH^-$ ?

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Explanation:

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How would you write the chemical equation for how the acetate buffer neutralizes excess acid, H3O+H_3O^+? How do you write the chemical equation for how the acetate buffer neutralizes excess base, OH−OH^-?

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How would you write the chemical equation for how the acetate buffer neutralizes excess acid, $H_{3} O^{+}$ $H_3O^+$ ? How do you write the chemical equation for how the acetate buffer neutralizes excess base, $O H^{-}$ $OH^-$ ?