Calculate pH for 0.10 mole NH3 dissolved in 2L of 0.050 M NH4NO3?

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Calculate pH for 0.10 mole NH3 dissolved in 2L of 0.050 M NH4NO3?

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Answer 1 · 2015-12-24T22:57:39

$pH = 9.26$ $"pH" = 9.26$

Explanation:

I will assume that you're not familiar with the Henderson - Hasselbalch equation, which allows you to calculate the pH or pOH of a buffer solution.

So, you're interested in finding the pH of a solution that contains $0.10$ $0.10$ moles of ammonia, ${NH}_{3}$ $"NH"_3$ , dissolved in $2 L$ $"2 L"$ of $0.050 M$ $"0.050 M"$ ammonium nitrate, ${NH}_{4} {NO}_{3}$ $"NH"_4"NO"_3$ solution.

As you know, ammonia is a weak base, which of course means that id does not ionize completely in aqueous solution to form ammonium ions, ${NH}_{4}^{+}$ $"NH"_4^(+)$ , its conjugate acid, and hydroxide ions, ${OH}^{-}$ $"OH"^(-)$ .

Instead, the following equilibrium will be established

${NH}_{3(aq]} + H_{2} O_{(l]} ⇌ {NH}_{4(aq]}^{+} + {OH}_{(aq]}^{-}$ $"NH"_text(3(aq]) + "H"_2"O"_text((l]) rightleftharpoons "NH"_text(4(aq])^(+) + "OH"_text((aq])^(-)$

Now, you're dissolving the ammonia in a solution that already contains ammonium ions, since ammonium nitrate, a soluble ionic compound, dissociates completely to form

${NH}_{4} {NO}_{3(aq]} \to {NH}_{4(aq]}^{+} + {NO}_{3(aq]}^{-}$ $"NH"_4"NO"_text(3(aq]) -> "NH"_text(4(aq])^(+) + "NO"_text(3(aq])^(-)$

Notice that ammonium nitrate dissociates in a $1 : 1$ $1:1$ mole ratio with the ammonium ions, which means that

$[{NH}_{4}^{+}] = [{NH}_{4} {NO}_{3}] = 0.050 M$ $["NH"_4^(+)] = ["NH"_4"NO"_3] = "0.050 M"$

The concentration of ammonia in this solution will be

$c = \frac{n}{V}$ $color(blue)(c = n/V)$

$[{NH}_{3}] = \frac{0.10 moles}{2 L} = 0.050 M$ $["NH"_3] = "0.10 moles"/"2 L" = "0.050 M"$

Use an ICE table to determine the equilibrium concentration of hydroxide ions in this solution

${NH}_{3(aq]} + H_{2} O_{(l]} ⇌ {NH}_{4(aq]}^{+} + {OH}_{(aq]}^{-}$ $" ""NH"_text(3(aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" " "NH"_text(4(aq])^(+) " "+" " "OH"_text((aq])^(-)$

$I 0.050 0.050 0$ $color(purple)("I")" " " "0.050" " " " " " " " " " " " " " " " "0.050" " " " " " " " "0$
$C (- x) (+ x) (+ x)$ $color(purple)("C")" "(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " "(+x)$
$E (0.050 - x) 0.050 + x x$ $color(purple)("E")" "(0.050-x)" " " " " " " " " " " "0.050+x" " " " " " "x$

By definition, the base dissociation constant, $K_{b}$ $K_b$ , will be equal to

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

$K_{b} = \frac{(0.050 + x) \cdot x}{0.050 - x}$ $K_b = ( (0.050+x) * x)/(0.050-x)$

You can find the value for $K_{b}$ $K_b$ here

http://www.bpc.edu/mathscience/chemistry/table_of_weak_bases.html

So, you have

$1.8 \cdot 10^{- 5} = \frac{(0.050 + x) \cdot x}{0.050 - x}$ $1.8 * 10^(-5) = ( (0.050+x) * x)/(0.050-x)$

Rearrange this equation to get

$x^{2} + (0.050 + 1.8 \cdot 10^{- 5}) \cdot x - 9 \cdot 10^{- 7} = 0$ $x^2 + (0.050 + 1.8 * 10^(-5)) * x - 9 * 10^(-7) = 0$

This quadratic solution will produce two solutions for $x$ $x$ , one positive and one negative. Since $x$ $x$ represents concentration, you can discard the negative value.

You will thus have

$x = 1.799 \cdot 10^{- 5}$ $x = 1.799 * 10^(-5)$

This means that you have

$[{OH}^{-}] = x = 1.799 \cdot 10^{- 5} M$ $["OH"^(-)] = x = 1.799 * 10^(-5)"M"$

The pOH of the solution will be

$pOH = - log ([{OH}^{-}])$ $color(blue)("pOH" = -log( ["OH"^(-)]))$

$pOH = - log (1.799 \cdot 10^{- 5}) = 4.74$ $"pOH" = - log(1.799 * 10^(-5)) = 4.74$

The pH of the solution will thus be

$pH = 14 - pOH$ $color(blue)("pH" = 14 - "pOH")$

$pH = 14 - 4.74 = 9.26$ $"pH" = 14 - 4.74 = color(green)(9.26)$

SIDE NOTE It is important to notice here that when you have equal concentrations of weak base and conjugate acid, the pOH of the solution will be equal to the $p K_{b}$ $pK_b$ of the weak base

$p K_{b} = - log (K_{b})$ $color(blue)(pK_b = - log(K_b))$

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

The Henderson - Hasselbalch equation for weak base / conjugate acid buffers looks like this

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

Notice that when

$[conjugate acid] = [weak base]$ $["conjugate acid"] = ["weak base"]$

you have

$pOH = p K_{b} + log (1) = p K_{b}$ $"pOH" = pK_b + log(1) = pK_b$

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Calculate pH for 0.10 mole NH3 dissolved in 2L of 0.050 M NH4NO3?

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