Question #52526

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Question #52526

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Answer 1 · 2017-04-07T00:02:35

Here's what I got.

Explanation:

For starters, you should know that ammonium chloride is a soluble ionic compound, which means that it dissociates completely when dissolved in water.

This means that you will have

${NH}_{4} {Cl}_{(a q)} \to {NH}_{4 (a q)}^{+} + {Cl}_{(a q)}^{-}$ $"NH"_ 4"Cl"_ ((aq)) -> "NH"_ (4(aq))^(+) + "Cl"_ ((aq))^(-)$

As you can see, the salt dissociates to produce ammonium cations and chloride anions in a $1 : 1$ $1:1$ mole ratio.

You can thus say that

$[{NH}_{4}^{+}] = 0.20 M$ $["NH"_4^(+)] = "0.20 M" " "$ and $[{Cl}^{-}] = 0.20 M$ $" " ["Cl"^(-)] = "0.20 M"$

Now, the ammonium cation will act as a weak acid in water and react to form hydronium cations and ammonia in a $1 : 1$ $1:1$ mole ratio.

${NH}_{4 (a q)}^{+} + H_{2} O_{(l)} ⇌ {NH}_{3 (a q)} + H_{3} O_{(a q)}^{+}$ $"NH"_ (4(aq))^(+) + "H"_ 2"O" _ ((l)) rightleftharpoons "NH"_ (3(aq)) + "H"_ 3"O"_ ((aq))^(+)$

Look up the acid dissociation constant for the ammonium cation

$K_{a} = 5.6 \cdot 10^{- 10}$ $K_a = 5.6 * 10^(-10)$

http://clas.sa.ucsb.edu/staff/Resource%20Folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

The $1 : 1$ $1:1$ mole ratio tells you that for every $1$ $1$ mole of ammonium cations that react with water, you will get $1$ $1$ mole of ammonia and $1$ $1$ mole of hydronium cations.

This means that if you take $x$ $x$ to be the concentration of ammonium cations that reacts with water, you can say that, at equilibrium, you will have

$[{NH}_{4}^{+}] = 0.20 M - x$ $["NH"_4^(+)] = "0.20 M" - x$

The concentration of ammonium cations will decrease by $x$ $x$ . Similarly, you will have

$[H_{3} O^{+}] = 0 M + x$ $["H"_3"O"^(+)] = "0 M" + x$

$[{NH}_{3}] = 0 M + x$ $["NH"_3] = "0 M" + x$

Mind you, you do have some hydronium cations present in water before you dissolve the salt, but their concentration is low enough to allow you to approximate it with $0$ $0$ .

By definition, the acid dissociation constant will be

$K_{a} = \frac{[{NH}_{3}] \cdot [H_{3} O^{+}]}{[{NH}_{4}^{+}]}$ $K_a = ( ["NH"_3] * ["H"_3"O"^(+)])/(["NH"_4^(+)])$

In your case, the will be equal to

$5.6 \cdot 10^{- 10} = \frac{x \cdot x}{0.20 - x} = \frac{x^{2}}{0.20 - x}$ $5.6 * 10^(-10) = (x * x)/(0.20 - x) = x^2/(0.20 - x)$

Because the acid dissociation constant has such a small value compared to the initial concentration of the ammonium cations, you can use the approximation

$0.20 - x \approx 0.20$ $0.20 - x ~~ 0.20$

This means that you will have

$5.6 \cdot 10^{- 10} = \frac{x^{2}}{0.20}$ $5.6 * 10^(-10) = x^2/0.20$

which gets you

$x = \sqrt{0.20 \cdot 5.6 \cdot 10^{- 10}} = 1.1 \cdot 10^{- 5}$ $x = sqrt(0.20 * 5.6 * 10^(-10)) = 1.1 * 10^(-5)$

You can thus say that your solution will contain

$[{NH}_{4}^{+}] = 0.20 - 1.1 \cdot 10^{- 5} = 0.199989 M$ $["NH"_4^(+)] = 0.20 - 1.1 * 10^(-5) ="0.199989 M"$

$[H_{3} O^{+}] = 1.1 \cdot 10^{- 5}$ $["H"_3"O"^(+)] = 1.1 * 10^(-5)$ $M$ $"M"$

$[{NH}_{3}] = 1.1 \cdot 10^{- 5}$ $["NH"_3] = 1.1 * 10^(-5)$ $M$ $"M"$

To find the percent dissociation, divide the concentration of the ammonium cations that dissociated to produce ions by the initial concentration of the ammonium cations and multiply the result by $100 %$ $100%$

$"% dissociation" = (1.1 * 10^(-5)color(red)(cancel(color(black)("M"))))/(0.20color(red)(cancel(color(black)("M")))) xx 100% = 0.0055%$

With the exception of the equilibrium concentration of ammonium cations, which I'll leave rounded to six sig figs for illustration purposes, all the values are rounded to two sig figs, the number of sig figs you have for the initial concentration of the salt.

As a bonus, you can calculate the concentration of hydroxide anions in this solution. To do that, use the fact that an aqueous solution at room temperature has

$["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-14)$ $"M"^2$

This will get you

$["OH"^(-)] = (10^(-14)"M"^color(red)(cancel(color(black)(2))))/(1.1 * 10^(-5)color(red)(cancel(color(black)("M")))) = 9.1 * 10^(-10)$ $"M"$

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Question #52526

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