What is the $pH$ $"pH"$ of a $1.5-M$ $"1.5-M"$ solution of ammonia? The dissociation constant of ammonia at ${25.0}^{\circ} C$ $25.0^@"C"$ is $1.80 \cdot 10^{- 5}$ $1.80 * 10^-5$ .

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What is the $pH$ $"pH"$ of a $1.5-M$ $"1.5-M"$ solution of ammonia? The dissociation constant of ammonia at ${25.0}^{\circ} C$ $25.0^@"C"$ is $1.80 \cdot 10^{- 5}$ $1.80 * 10^-5$ .

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Answer 1 · 2018-04-16T00:36:48

$pH = 11.72$ $"pH" = 11.72$

Explanation:

As you know, ammonia acts as a weak base in aqueous solution, so right from the start, you should expect the $pH$ $"pH"$ of the solution to be $> 7$ $> 7$ .

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

The ratio that exists between the equilibrium concentrations of the ammonium cations and of the hydroxide anions and the equilibrium concentration of ammonia is given by the base dissociation constant, $K_{b}$ $K_b$ .

$K_{b} = \frac{[{NH}_{4}^{+}] \cdot [{OH}^{-}]}{[{NH}_{3}]}$ $K_b = (["NH"_4^(+)] * ["OH"^(-)])/(["NH"_3])$

Now, ammonia will only partially ionize to produce ammonium cations and hydroxide anions. If you take $x$ $x$ $M$ $"M"$ to be the concentration of ammonia that ionizes, you can say that, at equilibrium, the solution will contain

$["NH"_4^(+)] = ["OH"^(-)] = x quad "M"$

This happens because every mole of ammonia that ionizes produces $1$ mole of ammonium cations and $1$ mole of hydroxide anions.

So if $x$ $"M"$ ionizes, you can expect the solution to contain $x$ $"M"$ of the two ions.

The solution will also contain

$["NH"_3] = (1.5 - x) quad "M"$

When $x$ $"M"$ ionizes, the initial concentration of ammonia will decrease by $x$ $"M"$ .

This means that the expression of the base dissociation constant will now take the form

$K_b = (x * x)/(1.5 - x)$

which is equal to

$1.80 * 10^(-5) = x^2/(1.5 - x)$

Notice that the value of the base dissociation constant is significantly smaller than the initial concentration of the base. This tells you that you can use the approximation

$1.5 -x ~~ 1.5$

because the concentration of ammonia that ionizes will be significantly lower than the initial concentration of the base, i.e. the ionization equilibrium will lie to the left.

You now have

$1.80 * 10^(-5) = x^2/1.5$

Rearrange and solve for $x$ to get

$x = sqrt(1.5 * 1.80 * 10^(-5)) = 0.005196$

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

$["OH"^(-)] = "0.005196 M"$

Now, an aqueous solution at $25^@"C"$ has

$"pH + pOH = 14"$

Since

$"pOH" = - log(["OH"^(-)])$

you can say that the $"pH"$ of the solution is given by

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

Plug in your value to find

$"pH" = 14 + log(0.005196) = color(darkgreen)(ul(color(black)(11.72)))$

The answer is rounded to two decimal places because you have two sig figs for the molarity of the solution.

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What is the $pH$ $"pH"$ of a $1.5-M$ $"1.5-M"$ solution of ammonia? The dissociation constant of ammonia at ${25.0}^{\circ} C$ $25.0^@"C"$ is $1.80 \cdot 10^{- 5}$ $1.80 * 10^-5$ .

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Explanation:

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What is the "pH"pH"pH" of a "1.5-M"1.5-M"1.5-M" solution of ammonia? The dissociation constant of ammonia at 25.0^@"C"25.0∘C25.0^@"C" is 1.80 * 10^-51.80⋅10−51.80 * 10^-5.

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Explanation:

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What is the $pH$ $"pH"$ of a $1.5-M$ $"1.5-M"$ solution of ammonia? The dissociation constant of ammonia at ${25.0}^{\circ} C$ $25.0^@"C"$ is $1.80 \cdot 10^{- 5}$ $1.80 * 10^-5$ .