Question #917b3

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Question #917b3

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Answer 1 · 2017-06-22T00:02:27

$pH = 11.20$ $"pH" = 11.20$

Explanation:

The idea here is that the cyanide anion, ${CN}^{-}$ $"CN"^(-)$ , will act as a weak base in aqueous solution, .e. it will react with water to form hydrocyanic acid, $HCN$ $"HCN"$ , and hydroxide anions, ${OH}^{-}$ $"OH"^(-)$ .

${CN}_{(a q)}^{-} + H_{2} O_{(l)} ⇌ {HCN}_{(a q)} + {OH}_{(a q)}^{-}$ $"CN"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HCN"_ ((aq)) + "OH"_ ((aq))^(-)$

So even without doing any calculations, you should expect the $pH$ $"pH"$ of the solution to come out $> 7$ $>7$ .

Now, look up the acid dissociation constant for hydrocyanic acid

$K_{a} = 6.17 \cdot 10^{- 10}$ $K_a = 6.17 * 10^(-10)$

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

Use the fact that an aqueous solution at room temperature has

$K_{a} \cdot K_{b} = K_{W} = 10^{- 14}$ $K_a * K_b = K_W = 10^(-14)$

to calculate the base dissociation constant of the cyanide anion.

$K_{b} = \frac{10^{- 14}}{6.17 \cdot 10^{- 10}} = 1.62 \cdot 10^{- 5}$ $K_b = (10^(-14))/(6.17 * 10^(-10)) = 1.62 * 10^(-5)$

Now, notice that every mole of cyanide anions that picks up a proton from water forms $1$ $1$ mole of hydrocyanic acid and $1$ $1$ mole of hydroxide anions.

If you take $x$ $x$ to be the equilibrium concentration of hydrocyanide acid and of hydroxide anions, you can say that

$[{CN}^{-}] = {[{CN}^{-}]}_{initial}^{-} - x$ $["CN"^(-)] = ["CN"^(-)]_ "initial" - x$

This means that in order to have a concentration of $x$ $x$ $M$ $"M"$ of hydrocyanide acid and of hydroxide anions, you need to consume $x$ $x$ $M$ $"M"$ of cyanide anions.

By definition, the base dissociation constant is equal to

$K_{b} = \frac{[HCN] \cdot [{OH}^{-}]}{[{CN}^{-}]}$ $K_b = (["HCN"] * ["OH"^(-)])/(["CN"^(-)])$

In your case, this will be equal to

$K_{b} = \frac{x \cdot x}{0.154 - x}$ $K_b = (x * x)/(0.154 - x)$

$1.62 \cdot 10^{- 5} = \frac{x^{2}}{0.154 - x}$ $1.62 * 10^(-5) = x^2/(0.154 - x)$

Now, notice that the base dissociation constant is considerably smaller than the initial concentration of the cyanide anions, so use the approximation

$0.154 - x \approx 0.154$ $0.154 - x ~~ 0.154$

This means that you have

$1.62 \cdot 10^{- 5} = \frac{x^{2}}{0.154}$ $1.62 * 10^(-5) = x^2/0.154$

Solve for $x$ $x$ to find

$x = \sqrt{0.154 \cdot 1.62 \cdot 10^{- 5}} = 1.58 \cdot 10^{- 3}$ $x = sqrt(0.154 * 1.62 * 10^(-5)) = 1.58 * 10^(-3)$

Since $x$ $x$ represents the equilibrium concentration of the hydroxide anions, you can say that you have

$[{OH}^{-}] = 1.58 \cdot 10^{- 3}$ $["OH"^(-)] = 1.58 * 10^(-3)$ $M$ $"M"$

An aqueous solution at room temperature has

$pH + pOH = 14$ $"pH + pOH = 14"$

so you can say that the $pH$ $"pH"$ of the solution is equal to

$pH = 14 - [- log ([{OH}^{-}])]$ $"pH" = 14 - [-log(["OH"^(-)])]$

$pH = 14 - [- log (1.58 \cdot 10^{- 3})]$ $"pH" = 14 - [-log(1.58 * 10^(-3))]$

$color(darkgreen)(ul(color(black)("pH" = 11.20)))$

I'll leave the answer rounded to two decimal places, but you can leave the answer rounded to three decimal places

$"pH" = 11.197$

because you have three sig figs for the initial concentration of the acid.

As predicted, the $"pH"$ of the solution is $>7$ , which is consistent with the fact that the cyanide anion acts as a weak base in aqueous solution.

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Question #917b3

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