Question #a3dd1

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Question #a3dd1

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Answer 1 · 2015-05-29T09:19:39

Your solution will have a pH equal to 4.0.

Since you're dealing with a weak acid, you can use an ICE table on its dissociation equilibrium to determine the equilibrium concentrations of the species involved in the reaction

$H A_{(a q)} + H_{2} O_{(l)} ⇌ H_{3} O_{(a q)}^{+} + A_{(a q)}^{-}$ $" "HA_((aq)) + H_2O_((l)) rightleftharpoons H_3O_((aq))^(+) + A_((aq))^(-)$
I.......1.........................................0.................0
C....(-x)......................................(+x).............(+x)
E.....1-x........................................x.................x

The acid dissociation constant, $K_{a}$ $K_a$ , will be equal to

$K_{a} = \frac{[H_{3} O^{+}] \cdot [A^{-}]}{[H A]} = \frac{x \cdot x}{1 - x} = \frac{x^{2}}{1 - x}$ $K_a = ([H_3O^(+)] * [A^(-)])/([HA]) = (x * x)/(1-x) = x^2/(1-x)$

Since $K_{a}$ $K_a$ has such a small value, you ca approximate (1-x) to be equal to 1, which would mean that

$K_{a} = \frac{x^{2}}{1} = x^{2} = 10^{- 8} \Rightarrow x = \sqrt{10^{- 8}} = 10^{- 4}$ $K_a = x^2/1 = x^2 = 10^(-8) => x = sqrt(10^(-8)) = 10^(-4)$

Since $x$ $x$ is equal to the equilibrium concentration of the hydronium ion, you'll have

$[H_{3} O^{+}] = 10^{- 4} M$ $[H_3O^(+)] = 10^(-4)"M"$

Therefore, the pH of the solution will be

$p H_{sol} = - log ([H_{3} O^{+}])$ $pH_"sol" = -log([H_3O^(+)])$

$p H_{sol} = - log (10^{- 4}) = 4.0$ $pH_"sol" = -log(10^(-4)) = color(green)(4.0)$

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