Question #7d9eb

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Question #7d9eb

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Answer 1 · 2015-10-31T22:50:43

$11.6 g$ $"11.6 g"$

Explanation:

The first thing to note here is that you need the value of ammonia's base dissociation constant, $K_{b}$ $K_b$ , which is listed as being equal to

$K_{b} = 1.8 \cdot 10^{- 5}$ $K_b = 1.8 * 10^(-5)$

Now, you can solve this relatively quickly by using the Henderson - Hasselbalch equation to find the $p O H$ $pOH$ of a buffer solution that contains ammonia, a weak base, and the ammonium ion, its conjugate acid

$p O H = p K_{b} + log (\frac{[conjugate acid]}{[weak base]})$ $color(blue)(pOH = pK_b + log( (["conjugate acid"])/(["weak base"])))$

Start by calculating the $p O H$ $pOH$ of the target solution

$p O H = 14 - p H$ $pOH = 14 - pH$

$p O H = 14 - 9.55 = 4.45$ $pOH = 14 - 9.55 = 4.45$

Use the base dissocaition constant to find $p K_{b}$ $pK_b$

$p K_{b} = - log ([{OH}^{-}])$ $pK_b = - log(["OH"^(-)])$

$p K_{b} = - log (1.8 \cdot 10^{- 5}) = 4.74$ $pK_b = - log(1.8 * 10^(-5)) = 4.74$

Plug these values into the Henderson - Hasselbalch equation to get

$4.45 = 4.74 + log (\frac{[{NH}_{4}^{+}]}{[{NH}_{3}]})$ $4.45 = 4.74 + log( (["NH"_4^(+)])/(["NH"_3]))$

$log (\frac{[{NH}_{4}^{+}]}{[{NH}_{3}]}) = 4.45 - 4.74$ $log( (["NH"_4^(+)])/(["NH"_3])) = 4.45 - 4.74$

This is equivalent to

$\frac{[{NH}_{4}^{+}]}{[{NH}_{3}]} = 10^{- 0.29}$ $(["NH"_4^(+)])/(["NH"_3]) = 10^(-0.29)$

$\frac{[{NH}_{4}^{+}]}{[{NH}_{3}]} = 0.5129$ $(["NH"_4^(+)])/(["NH"_3]) = 0.5129$

This means that the ratio between the concentration of the conjugate acid, which in your case will be provided by the ammonium chloride, and the concentration of the base must be equal to $0.7483$ $0.7483$ .

The concentration of the conjugate acid will thus be

$[{NH}_{4}^{+}] = [{NH}_{3}] \cdot 0.5129$ $["NH"_4^(+)] = ["NH"_3] * 0.5129$

. $[{NH}_{4}^{+}] = 0.160 M \cdot 0.5129 = 0.08206 M$ $["NH"_4^(+)] = "0.160 M" * 0.5129 = "0.08206 M"$

This means that you will need to provide the solution with

$c = \frac{n}{V} \Rightarrow n = c \cdot V$ $c = n/V implies n = c * V$

$n_{ammonium} = 0.08206 M \cdot 2.65 L = {0.2175 moles NH}_{4}^{+}$ $n_"ammonium" = "0.08206 M" * "2.65 L" = "0.2175 moles NH"_4^(+)$

SInce ammonium chloride dissociates in aqueous solution to give

${NH}_{4} {Cl}_{(aq]} \to {NH}_{4(aq]}^{+} + {Cl}_{(aq]}^{-}$ $"NH"_4"Cl"_text((aq]) -> "NH"_text(4(aq])^(+) + "Cl"_text((aq])^(-)$

it follows that the number of moles of ammonium chloride will be equal to the number of moles of ammonium ions.

$n_{N H_{4} C l} = n_{N H_{4}^{+}} = 0.2175 moles$ $n_(NH_4Cl) = n_(NH_4^(+)) = "0.2175 moles"$

To find how many grams of ammonium chloride would contain this many moles, use the compound's molar mass

$0.2175color(red)(cancel(color(black)("moles"))) * "53.49 g"/(1color(red)(cancel(color(black)("mole")))) = "11.634 g NH"_4"Cl"$

Rounded to three sig figs, the answer will be

$m_(NH_4Cl) = color(green)("11.6 g")$

SIDE NOTE I recommend double-checking the result by using an ICE table to find the cocnentration of ammonium ions. It would be good practice.

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Question #7d9eb

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