Question #156ba

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Question #156ba

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Answer 1 · 2017-04-25T02:31:35

pH = 5.13

Explanation:

We will use the $Henderson-Hasselbach$ $"Henderson-Hasselbach"$ equation to find the pH of this buffer solution

$a a a a a a a a$ $color(white)(aaaaaaaa)$ $p H = p K a + log [\frac{conjugate base}{weak acid}]$ $color(magenta)(pH = pKa + log[("conjugate base")/("weak acid"]]$

$Step 1$ $"Step 1"$
Find the equilibrium concentrations of the $acetic acid$ $"acetic acid"$ (weak acid) and its conjugate base, $sodium acetate$ $"sodium acetate"$ . Find moles of each solute from their initial volume and initial concentration. Use total volume as $0.034 L$ $0.034" L"$ .

$Acetic acid:$ $color(orange)"Acetic acid:"$

$(0.010 L) \cdot (1.0 M) = 0.01 moles$ $(0.010" L") * (1.0" M") = 0.01" moles"$
$\frac{0.01 moles}{0.034 L} = 0.29 M$ $(0.01" moles")/(0.034" L") = 0.29" M"$

$Sodium acetate:$ $color(green)"Sodium acetate:"$

$(0.024 L) \cdot (1.0 M) = 0.024 moles$ $(0.024" L") * (1.0" M") = 0.024" moles"$
$\frac{0.024 moles}{0.034 L} = 0.71 M$ $(0.024" moles")/(0.034" L") =0.71" M"$

$Step 2$ $"Step 2"$
Look up the $K_{a}$ $K_(a)$ of $acetic acid$ $"acetic acid"$ from your chemistry book or the Web.
$a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaa)$ $K_{a} = 1.8 \cdot 10^{- 5}$ $K_(a) = 1.8 * 10^-5$
(source:preparatorychemistry.com/Bishop_weak_acid_Equilibrium.htm)

To find the $p K a$ $pKa$ of acetic acid, take the negative log of the $K_{a}$ $K_(a)$

$p K a = - log (1.8 \cdot 10^{- 5})$ $pKa = -log(1.8*10^-5)$
$p K a = 4.74$ $pKa = 4.74$

$Step 3$ $"Step 3"$
Plugin and solve

$p H = p K a + log [\frac{conjugate base}{acid}]$ $pH = pKa + log[("conjugate base")/("acid")]$
$pH = 4.74 + log[(0.71 cancelM)/(0.29 cancelM")]$
$pH = 4.74 + log[2.45]$
$pH = 4.74 + 0.39$
$pH = 5.13$

You can check your answer by seeing that you clearly have more than $"2x"$ the amount of the $"conjugate base"$ than the $"weak acid"$ so your answer should reflect a higher $"pH"$ .

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Question #156ba

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