How do you calculate the pH of acetic acid?

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How do you calculate the pH of acetic acid?

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Answer 1 · 2016-06-24T01:06:17

Here's how you can do that.

Explanation:

Acetic acid, ${CH}_{3} COOH$ $"CH"_3"COOH"$ , is a weak acid, meaning that it partially ionizes in aqueous solution to form hydronium cations, $H_{3} O^{+}$ $"H"_3"O"^(+)$ , and acetate anions, ${CH}_{3} {COO}^{-}$ $"CH"_3"COO"^(-)$ .

${CH}_{3} COO H_{(a q)} + H_{2} O_{(l)} ⇌ H_{3} O_{(a q)}^{+} + {CH}_{3} {COO}_{(a q)}^{-}$ $"CH"_ 3"COO"color(red)("H")_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(color(red)(+)) + "CH"_ 3"COO"_ ((aq))^(-)$

The position of the ionization equilibrium is given by the acid dissociation constant, $K_{a}$ $K_a$ , which for acetic acid is equal to

$K_{a} = 1.8 \cdot 10^{- 5}$ $K_a = 1.8 * 10^(-5)$

http://www.bpc.edu/mathscience/chemistry/table_of_monoprotic_acids.html

Now, let's assume that you want to find the pH of a solution of acetic acid that has a concentration of $c$ $c$ . According to the balanced chemical equation that describes the ionization of the acid, every mole of acetic acid that ionizes will produce

one mole of hydronium cations

one mole of acetate anions

If you take $x$ $x$ to be the concentration of acetic acid that ionizes, you can find the equilibrium concentration of the hydronium cations by using an ICE table

${CH}_{3} COO H_{(a q)} + H_{2} O_{(l)} ⇌ H_{3} O_{(a q)}^{+} + {CH}_{3} {COO}_{(a q)}^{-}$ $"CH"_ 3"COO"color(red)("H")_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(color(red)(+)) + "CH"_ 3"COO"_ ((aq))^(-)$

$I a a a a a a a c a a a a a a a a a a a a a a a a a a a 0 a a a a a a a a a a a a 0$ $color(purple)("I")color(white)(aaaaaaacolor(black)(c)aaaaaaaaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaaacolor(black)(0)$
$C a a a a (- x) a a a a a a a a a a a a a a a a (+ x) a a a a a a a a (+ x)$ $color(purple)("C")color(white)(aaaacolor(black)((-x))aaaaaaaaaaaaaaaacolor(black)((+x))aaaaaaaacolor(black)((+x))$
$E a a a a a c - x a a a a a a a a a a a a a a a a a a x a a a a a a a a a a a x$ $color(purple)("E")color(white)(aaaaacolor(black)(c-x)aaaaaaaaaaaaaaaaaacolor(black)(x)aaaaaaaaaaacolor(black)(x)$

The acid dissociation constant will be equal to

$K_{a} = \frac{[H_{3} O^{+}] \cdot [{CH}_{3} {COO}^{-}]}{[{CH}_{3} COOH]}$ $K_a = (["H"_3"O"^(+)] * ["CH"_3"COO"^(-)])/(["CH"_3"COOH"])$

This will be equivalent to

$K_{s p} = \frac{x \cdot x}{c - x} = \frac{x^{2}}{c - x}$ $K_(sp) = (x * x)/(c-x) = x^2/(c-x)$

Now, as long as the initial concentration of the acetic acid, $c$ $c$ , is significantly higher than the $K_{s p}$ $K_(sp)$ of the acid, you can use the approximation

$c - x \approx c \to$ $c - x ~~ c ->$ valid when $color(red)(ul(color(black)(c " >> " K_(sp)))$

In this case, the equation becomes

$K_(sp) = x^2/c$

which gives you

$x = sqrt(c * K_(sp))$

Since $x$ represents the equilibrium concentration of hydronium cations, you will have

$["H"_3"O"^(+)] = sqrt(c * K_(sp))$

Now, the pH of the solution is given by

$color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))$

Combine these two equations to get

$color(green)(|bar(ul(color(white)(a/a)color(black)("pH" = - log( sqrt(c * K_(sp)))color(white)(a/a)|)))$

For example, the pH of a $"0.050 M"$ acetic acid solution will be

$"pH" = - log( 0.050 * 1.8 * 10^(-5))$

$"pH" = 3.02$

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How do you calculate the pH of acetic acid?

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