Question #27fca

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Question #27fca

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Answer 1 · 2016-02-09T00:25:37

$pH = 3.27$ $"pH" = 3.27$

Explanation:

Before doing any calculation, try to predict what you expect the concentrations of $H_{2} A$ $"H"_2"A"$ and $A^{2 -}$ $"A"^(2-)$ to be relative to each other.

Notice that he acid dissociation constant for the first ionization, $K_{a 1}$ $K_(a1$ , if about three orders of magnitude larger than the acid dissociation constant for the second ionization, $K_{a 2}$ $K_(a2)$ .

Moreover, the value of $K_{a 2}$ $K_(a2)$ is very, very small to begin with. This means that you can expect to have a very, very small concentration of $A^{2 -}$ $"A"^(2-)$ when equilibrium sets in.

In addition to this, you an expect almost all of the hydronium ions, $H_{3} O^{+}$ $"H"_3"O"^(+)$ , present in solution at equilibrium to come from the first ionization of the acid.

With this in mind, set up an ICE table for the first ionization of the acid

$H_{2} A_{(aq]} + H_{2} O_{(l]} ⇌ H_{3} O_{(aq]}^{+} + {HA}_{(aq]}^{-}$ $" " "H"_2"A"_text((aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" " "H"_3"O"_text((aq])^(+) " "+" " "HA"_text((aq])^(-)$

$I 0.0800 00$ $color(purple)("I")" " " "0.0800" " " " " " " " " " " " " " " "0" " " " " " " " " " " "0$
$C (- x) (+ x) (+ x)$ $color(purple)("C")" " " "(-x)" " " " " " " " " " " " " "(+x)" " " " " " " "(+x)$
$E 0.0800 - x x x$ $color(purple)("E")" "0.0800-x" " " " " " " " " " " " " "x" " " " " " " " " " " "x$

By definition, the acid dissociation constant will be equal to

$K_{a 1} = \frac{[H_{3} O^{+}] \cdot [{HA}^{-}]}{[H_{2} A]} = 3.6 \cdot 10^{- 6}$ $K_(a1) = (["H"_3"O"^(+)] * ["HA"^(-)])/(["H"_2"A"]) = 3.6 * 10^(-6)$

This is equivalent to

$3.6 \cdot 10^{- 6} = \frac{x \cdot x}{0.0800 - x}$ $3.6 * 10^(-6) = (x * x)/(0.0800 - x)$

Since $K_{a 1}$ $K_(a1)$ is so small compared with $0.0800$ $0.0800$ , you can use the approximation

$0.0800 - x \approx 0.0800$ $0.0800 - x ~~ 0.0800$

This will get you

$3.6 \cdot 10^{- 6} = \frac{x^{2}}{0.0800}$ $3.6 * 10^(-6) = x^2/0.0800$

$x = \sqrt{0.0800 \cdot 3.6 \cdot 10^{- 6}} \Rightarrow x = 5.37 \cdot 10^{- 4}$ $x = sqrt(0.0800 * 3.6 * 10^(-6)) implies x = 5.37 * 10^(-4)$

Since $x$ $x$ represents the concentration of hydronium ions produced by the first ionization of the acid and the concentration of ${HA}^{-}$ $"HA"^(-)$ , you will have

$[H_{2} A] = 0.0800 M - 5.37 \cdot 10^{- 4} M = 0.079466 M$ $["H"_2"A"] = "0.0800 M" - 5.37 * 10^(-4)"M" = "0.079466 M"$

$[H_{3} O^{+}] = 5.37 \cdot 10^{- 4} M$ $["H"_3"O"^(+)] = 5.37 * 10^(-4)"M"$

$[{HA}^{-}] = 5.37 \cdot 10^{- 4} M$ $["HA"^(-)] = 5.37 * 10^(-4)"M"$

Now, you could skip the second ionization altogether, since you have such a small starting concentration for ${HA}^{-}$ $"HA"^(-)$ , but let's go through with it just to test if our initial estimations were correct.

Use these concentrations to set up a second ICE table, this time of the second ionization of the acid

${HA}_{(aq]}^{-} + H_{2} O_{(l]} ⇌ H_{3} O_{(aq]}^{+} + A_{(aq]}^{2 -}$ $" " "HA"_text((aq])^(-) " "+ "H"_2"O"_text((l]) " "rightleftharpoons" " "H"_3"O"_text((aq])^(+) " "+" " "A"_text((aq])^(2-)$

$I 5.37 \cdot 10^{- 4} 5.37 \cdot 10^{- 4} 0$ $color(purple)("I")" "5.37 * 10^(-4)" " " " " " " " " " " "5.37 * 10^(-4)" " " " " " " " "0$
$C (- x) (+ x) (+ x)$ $color(purple)("C")" "(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " " " "(+x)$
$E 5.37 \cdot 10^{- 4} - x 5.37 \cdot 10^{- 4} + x x$ $color(purple)("E")" "5.37 * 10^(-4)-x" " " " " " " "5.37 * 10^(-4)+x" " " " " "x$

This time, the acid dissociation constant will be

$K_{a 2} = \frac{[H_{3} O^{+}] \cdot [A^{2 -}]}{[{HA}^{-}]} = 6.8 \cdot 10^{- 9}$ $K_(a2) = (["H"_3"O"^(+)] * ["A"^(2-)])/(["HA"^(-)]) = 6.8 * 10^(-9)$

This will be equivalent to

$6.8 \cdot 10^{- 9} = \frac{(5.37 \cdot 10^{- 4} + x) \cdot x}{5.37 \cdot 10^{- 4} - x}$ $6.8 * 10^(-9) = ((5.37 * 10^(-4) + x) * x)/(5.37 * 10^(-4) - x)$

Once again, because $K_{a 2}$ $K_(a2)$ has such a small value compared with $5.37 \cdot 10^{- 4}$ $5.37 * 10^(-4)$ , you can say that

$5.37 \cdot 10^{- 4} + x \approx 5.37 \cdot 10^{- 4}$ $5.37 * 10^(-4) + x ~~ 5.37 * 10^(-4)$

$5.37 \cdot 10^{- 4} - x \approx 5.37 \cdot 10^{- 4}$ $5.37 * 10^(-4) - x ~~ 5.37 * 10^(-4)$

This will give you

$6.8 * 10^(-9) = (color(red)(cancel(color(black)(5.37 * 10^(-4)))) * x)/color(red)(cancel(color(black)(5.37 * 10^(-4)))) = x$

Therefore, the equilibrium concentration of hydronium ions will be

$["H"_3"O"^(+)] = 5.37 * 10^(-4) + 6.98 * 10^(-9) ~~ 5.37 * 10^(-4)"M"$

As predicted, the first ionization determined the equilibrium concentration of hydronium ions.

The pH of the solution will be

$color(blue)("pH" = - log(["H"_3"O"^(+)]))$

$"pH" = - log(5.37 * 10^(-4)) = color(green)(3.27)$

The concentrations of $"H"_2"A"$ and $"A"^(2-)$ will be

$["H"_2"A"] = color(green)("0.0795 M")$

$["A"^(2-)] = color(green)(6.80 * 10^(-9)"M")$

I'll leave these rounded to three sig figs.

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