Question #26e0c

Apr 24, 2015

The answer is (3), that particular mixture of ammonium sulfate and sodium hydroxide will show buffer action.

Here's how the problem looks like

A buffer solution will consist of a weak acid and its conjugate base, or of a weak base and its conjugate acid

For a solution to exhibit buffer action, it must be able to resist significant pH changes when small quantities of a strong acid or of a strong base are added to it.

At first, neither of the proposed solutions look like they could act as a buffer, but one of them can indeed do so.

Take a look at the reaction between ammonium sulfate and sodium hydroxide

${\left(N {H}_{4}\right)}_{2} S {O}_{4} + \textcolor{red}{2} N a O H \to N {a}_{2} S {O}_{4} + 2 N {H}_{3} + 2 {H}_{2} O$

The net ionic equation will be

$N {H}_{4}^{+} + O {H}^{-} \to N {H}_{3} + {H}_{2} O$

Notice that this reaction produces ammonia, which is a weak base. depending on how many moles of each reactant were added together, the solution can now contain ammonia, a weak base, and ammonium ion, its conjugate acid.

Use the molarities of the two solutions to determine how many moles you add together. For option (3) you'd get

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{{\left(N {H}_{4}\right)}_{2} S {O}_{4}} = 10. \cdot {10}^{- 3} \text{L" * "1 M" = "0.01 moles }$ ${\left(N {H}_{4}\right)}_{2} S {O}_{4}$

${n}_{N a O H} = 10. \cdot {10}^{- 3} \text{L" * "1 M" = "0.01 moles }$ $N a O H$

Since 1 mole of ammonium sulfate produces 2 moles of ammonium ions in solution, you'll get

${n}_{N {H}_{4}^{+}} = 0.01 \cdot 2 = \text{0.02 moles }$ $N {H}_{4}^{+}$

Look at the net ionic equation. You have a $1 : 1$ mole ratio between the ammonium and hydroxide ions; this means that all of the hydroxide will be consumed, and you'll be left with

${n}_{N {H}_{4}^{+}} = 0.02 - 0.01 = \text{0.01 moles }$ $N {H}_{4}^{+}$

At the same time, the reaction will produce 0.01 moles of ammonia, so you'll now have both a weak base and its conjugate acid in solution $\to$ can act as a buffer.

If you look at option (4), you'll notice that you add twice as many moles of sodium hydroxide, which implies that

${n}_{O {H}^{-}} = \text{0.02 moles}$

${n}_{N {H}_{4}^{+}} = \text{0.02 moles}$ $\to$ since you add the same amount of ammonium sulfate as you did at example (3).

Both the reactants will be consumed in the reaction, so the resulting solution will only contain ammonia, a weak base, and not its conjugate acid.

Options (1) and (2) cannot exhibit buffer action because they lack one of the two components needed (a weak base for option 1 and a weak acid for option 2).