AP Chem 3.8: Representations of Solutions

Zooming in on a solution

The AP exam rarely asks only for a concentration. It often shows several particle diagrams of a solution and asks which is correct, or asks you to draw what a mixture looks like. AP topic 3.8 asks you to turn concentration, dissociation, and precipitation into particle diagrams.

Four rules for drawing solutions

Strong electrolytes dissociate completely: the ions are separate; don’t draw “NaCl molecules” in pairs.
The ratio matches the formula: in $\ce{MgCl2}$ solution there are twice as many $\ce{Cl-}$ as $\ce{Mg^2+}$ ; in $\ce{Na2SO4}$ solution twice as many $\ce{Na+}$ as $\ce{SO4^2-}$ .
Water molecules point the right way: around a cation, the O end of water (δ−) faces the ion; around an anion, the H end (δ+) faces the ion (the ion–dipole forces of U3.1).
Molecular compounds don’t dissociate: in glucose or ethanol solution, draw whole molecules, not separate atoms, and no ions.

Concentration is proportional to the number of particles per unit volume: the more crowded the drawing, the higher the concentration. On dilution the number of particles stays the same, spread through a larger volume. When equal volumes of two solutions are mixed, an ion that comes from only one of them ends up at half its original concentration.

After a precipitation reaction

When $\ce{AgNO3}$ solution and NaCl solution are mixed, $\ce{Ag+}$ and $\ce{Cl-}$ combine to form insoluble AgCl:

\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}

$\ce{Na+}$ and $\ce{NO3-}$ don’t take part; they are spectator ions旁观离子. The particle diagram after the reaction should show:

solid AgCl at the bottom, whose $\ce{Ag+}$ and $\ce{Cl-}$ are no longer free ions in solution;
all of the spectator ions ( $\ce{Na+}$ and $\ce{NO3-}$ ) in solution;
if one reactant was in excess, the leftover ions of that reactant ( $\ce{Ag+}$ or $\ce{Cl-}$ ) in solution too; if the amounts matched exactly, none.

First see what a dissolved salt looks like, then mix two solutions and change the amount of B to see the precipitate and the ions left over:

Try it: particle diagrams of solutions

Example 1: A particle diagram of Na₂SO₄ solution

Describe a particle diagram of $\ce{Na2SO4}$ solution: if 2 formula units have dissolved, how many of which ions should be drawn, and how are the water molecules arranged?

Step-by-step solution

Write the dissociation. $\ce{Na2SO4}$ is a strong electrolyte and dissociates completely:
$\ce{Na2SO4 -> 2Na+ + SO4^2-}$
Count the ions. 2 formula units give 4 $\ce{Na+}$ and 2 $\ce{SO4^2-}$ , drawn separately, not as “ $\ce{Na2SO4}$ molecules.”
Orient the water molecules. Around $\ce{Na+}$ , water molecules point their partially negative O toward the ion; around $\ce{SO4^2-}$ , they point their partially positive H atoms toward it.
Check the charge.
$4 \times (+1) + 2 \times (-2) = 0$
The solution is neutral, so the diagram is consistent.

Example 2: One reactant in excess

A solution containing 3 formula units of $\ce{AgNO3}$ is mixed with one containing 2 formula units of NaCl. After the reaction, what is at the bottom of the beaker and what is in solution?

Step-by-step solution

List the ions before mixing. 3 $\ce{Ag+}$ , 3 $\ce{NO3-}$ , 2 $\ce{Na+}$ , 2 $\ce{Cl-}$ .
Write the net ionic equation and find the limiting ion.
$\ce{Ag+ + Cl- -> AgCl(s)}$
$\ce{Ag+}$ and $\ce{Cl-}$ react 1 : 1, and there are only 2 $\ce{Cl-}$ , so $\ce{Cl-}$ is limiting: 2 AgCl form and settle to the bottom.
Tabulate before and after.

Particle Before After
$\ce{Ag+}$ 3 1
$\ce{Cl-}$ 2 0
AgCl (bottom) 0 2
$\ce{Na+}$ 2 2
$\ce{NO3-}$ 3 3

Left in solution: 1 excess $\ce{Ag+}$ and the spectator ions $\ce{Na+}$ and $\ce{NO3-}$ .
Check the charge. In solution, positive charge 1 + 2 = 3 and negative charge 3: neutral.

Particle	Before	After
$\ce{Ag+}$	3	1
$\ce{Cl-}$	2	0
AgCl (bottom)	0	2
$\ce{Na+}$	2	2
$\ce{NO3-}$	3	3

Practice: is this description of a particle diagram correct?

KBr solution: equal numbers of $\ce{K+}$ and $\ce{Br-}$ , all separate
$\ce{MgCl2}$ solution: each $\ce{Mg^2+}$ with two $\ce{Cl-}$ stuck to it in clusters
Around $\ce{Na+}$ , water molecules point their H atoms toward the ion
Around $\ce{Cl-}$ , water molecules point their H atoms toward the ion
$\ce{Na2SO4}$ solution: 4 $\ce{Na+}$ and 2 $\ce{SO4^2-}$
$\ce{AlCl3}$ solution: equal numbers of $\ce{Al^3+}$ and $\ce{Cl-}$
Glucose solution: whole glucose molecules
Glucose solution: each molecule split into C, H, and O atoms
$\ce{AgNO3}$ and NaCl in exactly matching amounts, after reacting: only $\ce{Na+}$ and $\ce{NO3-}$ in solution, AgCl at the bottom
$\ce{AgNO3}$ and NaCl in exactly matching amounts, after reacting: $\ce{Ag+}$ and $\ce{Cl-}$ still in solution
After a KCl solution is mixed with an equal volume of water, the number of $\ce{K+}$ per unit volume is halved

Common mistakes

Mistake 1: “An ionic compound dissolved in water exists as molecules such as NaCl.” Ionic compounds dissociate completely when they dissolve; the cations and anions are separated and surrounded by water molecules.

Mistake 2: “Water molecules point their H ends toward cations.” A cation is positive, so it attracts the partially negative O end of water; the H ends point toward anions.

Mistake 3: “After a precipitate forms, its ions still move freely in solution.” The precipitate is a solid with its ions fixed in a lattice. Only the spectator ions and any ion in excess remain in solution.

Quick check

Which diagram best represents a solution of $\ce{MgCl2}$ ? (Most water molecules are left out.)

A solution containing 3 formula units of $\ce{AgNO3}$ reacts exactly with one containing 3 formula units of $\ce{NaCl}$ . Which diagram best represents the mixture after the reaction?

A particle diagram of NaCl solution shows 4 $\ce{Na+}$ and 4 $\ce{Cl-}$ , each ion representing 0.050 M. After an equal volume of water is added, what is $[\ce{Na+}]$ ?

Equal volumes of 0.20 M KCl and 0.20 M $\ce{NaNO3}$ are mixed (no reaction). What are $[\ce{K+}]$ and $[\ce{NO3-}]$ in the mixture?

Solutions of $\ce{BaCl2}$ and $\ce{Na2SO4}$ react in exactly matching amounts. Which ions are present in significant amounts in the solution afterward?

A solution containing 2 formula units of $\ce{BaCl2}$ is mixed with one containing 4 formula units of $\ce{Na2SO4}$ . Which ions are in solution after the reaction?

AP-style free response

When solutions of $\ce{Pb(NO3)2}$ and KI are mixed, a yellow precipitate of $\ce{PbI2}$ forms.

(a) Write the net ionic equation for the reaction.

(b) A solution containing 2 formula units of $\ce{Pb(NO3)2}$ is mixed with one containing 6 formula units of KI. After the reaction, how many of which particles are at the bottom and in solution?

(c) Describe the particle diagram of the mixture in (b) after the reaction.

(d) 50.0 mL of 0.100 M $\ce{Pb(NO3)2}$ is mixed with 50.0 mL of 0.300 M KI. Find $[\ce{K+}]$ and $[\ce{I-}]$ in the solution after the reaction.

Model answer and scoring

(a) $\ce{Pb^2+(aq) + 2I-(aq) -> PbI2(s)}$

(b) 2 $\ce{Pb^2+}$ need 4 $\ce{I-}$ , forming 2 $\ce{PbI2}$ at the bottom. In solution: 2 $\ce{I-}$ left over (in excess), plus the spectator ions, 6 $\ce{K+}$ and 4 $\ce{NO3-}$ ; no $\ce{Pb^2+}$ .

(c) 2 units of solid $\ce{PbI2}$ at the bottom; in solution, 6 $\ce{K+}$ , 4 $\ce{NO3-}$ , and 2 $\ce{I-}$ , separate and evenly spread; the charges balance (6 = 4 + 2).

(d) n( $\ce{K+}$ ) = 0.0500 × 0.300 = 0.0150 mol, which doesn’t react, so $[\ce{K+}]$ = 0.0150 ÷ 0.100 = 0.150 M. n( $\ce{Pb^2+}$ ) = 0.00500 mol, which uses up 0.0100 mol of $\ce{I-}$ ; of the 0.0150 mol of $\ce{I-}$ , 0.00500 mol remains, so $[\ce{I-}]$ = 0.0500 M.

Scoring: (a) balanced net ionic equation; (b) 2 units of precipitate, with 2 $\ce{I-}$ , 6 $\ce{K+}$ , and 4 $\ce{NO3-}$ left; (c) precipitate at the bottom, ions separate, charge balanced; (d) 0.150 M and 0.0500 M, using 100.0 mL.

Summary

Solutions of strong electrolytes: ions separate, in the ratio of the formula; water’s O faces cations and its H atoms face anions. Draw whole molecules for nonelectrolytes.
More particles per unit volume means higher concentration; on dilution or mixing equal volumes, the number of particles stays the same and the volume grows.
After a precipitation: the precipitate is at the bottom, and only spectator ions and any ion in excess remain in solution, with the charges balanced.

AP tip: When choosing a particle diagram, check each point: Are the ions separate? Is the ratio right? Do the water molecules point the right way? Do the charges balance? After a reaction, are any spectator ions missing, or are ions that have precipitated still drawn in solution?

Representations of Solutions溶液的粒子图表示

Learning goals

Before you start

Zooming in on a solution

Four rules for drawing solutions

After a precipitation reaction

Example 1: A particle diagram of Na₂SO₄ solution

Example 2: One reactant in excess

Common mistakes

Quick check

AP-style free response

Summary

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