AP Chem 3.7: Solutions and Mixtures

A solution is a homogeneous mixture

A solution is a homogeneous mixture: the particles of the solute (molecules or ions) are spread evenly through the solvent, so any small portion has the same composition. Chapter 2.9 covered molarity and dilution; AP topic 3.7 asks you to calculate the concentration of each ion in a solution fluently, describe how to prepare solutions correctly, and compare how well solutions conduct.

Molarity and ion concentrations

M = \frac{\text{mol solute}}{\text{L solution}}

Note that the denominator is the volume of solution, not the volume of solvent.

A strong electrolyte dissociates completely in water, so the concentration of each ion equals the concentration of the solute times the number of that ion in the formula:

0.20 M $\ce{MgCl2}$ : $[\ce{Mg^2+}]$ = 0.20 M and $[\ce{Cl-}]$ = 0.40 M.
0.15 M $\ce{Na2SO4}$ : $[\ce{Na+}]$ = 0.30 M and $[\ce{SO4^2-}]$ = 0.15 M.
Working backward, if $[\ce{NO3-}]$ = 0.30 M in an $\ce{Al(NO3)3}$ solution, then $[\ce{Al^3+}]$ = 0.10 M.

Mixing two solutions

When two solutions that don’t react are mixed (treating volumes as additive):

Find the moles of the ion in each solution: n = M × V × its subscript.
Add the moles of the same ion together.
Divide by the total volume after mixing.

If equal volumes are mixed and an ion comes from only one of the solutions, that ion’s concentration is halved. If the ions of the two solutions form a precipitate, the reaction must be accounted for first (U3.8).

Report results to the significant figures the data allow: 0.0350 mol of solute in 250.0 mL of solution is 0.140 M (3 significant figures).

Try it: find the concentration of each ion, then mix two solutions

Solution AConcentration (M)Volume (mL)

Solution BConcentration (M)Volume (mL)

Preparing and diluting solutions

To prepare a solution of known concentration from a solid, use a volumetric flask容量瓶: dissolve the solute, then add water up to the calibration mark so that the total volume of solution is exactly the volume you need. Don’t add the solute to “1.00 L of water”; the solution’s volume would then not be 1.00 L.

Put the steps for preparing 250.0 mL of 0.100 M Na₂CO₃ in order

Steps to place (click to add)

Your order (click a step to send it back)

To measure out a volume of solution accurately, use a volumetric pipet移液管 (or a buret); a graduated cylinder is only good for approximate volumes, and the markings on a beaker are rough guides.

On dilution, the moles of solute don’t change:

M_1V_1 = M_2V_2

Be careful: V₂ is the total volume of the diluted solution, not the volume of water added. To make 100.0 mL of 0.500 M solution from 2.00 M stock, take 25.0 mL of stock and dilute it with water to 100.0 mL in a volumetric flask, rather than “adding 100.0 mL of water”.

Electrolytes and conductivity

A solution conducts electricity through its mobile ions. At the same concentration:

Strong electrolytes (such as NaCl and $\ce{Na2SO4}$ ) dissociate completely, give many ions, and conduct well. The more ions each formula unit releases, the higher the total ion concentration: 0.10 M $\ce{Na2SO4}$ (0.30 M ions in all) conducts better than 0.10 M NaCl (0.20 M).
Weak electrolytes (such as acetic acid, $\ce{CH3COOH}$ ) ionize only slightly, give few ions, and conduct poorly.
Nonelectrolytes (such as glucose and ethanol) dissolve as molecules and don’t conduct.

Example 1: [Cl⁻] after mixing

50.0 mL of 0.200 M $\ce{MgCl2}$ is mixed with 25.0 mL of 0.100 M KCl (assume the volumes add). Find [Cl⁻] in the mixture.

Step-by-step solution

Find the Cl⁻ from each solution. With volumes in mL, mL × M gives millimoles. Each $\ce{MgCl2}$ gives 2 Cl⁻ and each KCl gives 1 (in mmol):
$\begin{aligned} n_1 &= 50.0 \times 0.200 \times 2 = 20.0 \\ n_2 &= 25.0 \times 0.100 \times 1 = 2.50 \end{aligned}$
Add them and divide by the total volume. The total volume is 50.0 + 25.0 = 75.0 mL, and mmol/mL is the same as mol/L:
$\begin{aligned} [\ce{Cl-}] &= \frac{20.0 + 2.50}{75.0} \\ &= \boxed{0.300\ \text{M}} \end{aligned}$
Check. [Mg²⁺] = 10.0 ÷ 75.0 = 0.133 M and [K⁺] = 2.50 ÷ 75.0 = 0.0333 M, so the positive charge is 2 × 0.133 + 0.0333 = 0.300 M, matching [Cl⁻]: charge balances.
Common mistakes. Forgetting that $\ce{MgCl2}$ has 2 Cl⁻, or simply adding the two concentrations.

Example 2: Preparing a dilute solution from a concentrated one

How many milliliters of 2.00 M NaCl are needed to make 250.0 mL of 0.100 M NaCl, and how is it done?

Step-by-step solution

The moles of solute don’t change on dilution.
$\begin{aligned} M_1V_1 &= M_2V_2 \\ V_1 &= \frac{M_2V_2}{M_1} = \frac{(0.100)(250.0)}{2.00} \\ &= \boxed{12.5\ \text{mL}} \end{aligned}$
Measure. Use a pipet (or buret) to deliver exactly 12.50 mL of the concentrated solution into a 250 mL volumetric flask.
Make up to the mark. Add distilled water up to the mark, stopper, and invert several times to mix.
Note. About 237.5 mL of water ends up being added, but you don’t measure the water; you fill to the mark. “Add 12.5 mL of stock to 250.0 mL of water” is wrong, because the total volume would exceed 250 mL.

Example 3: Preparing a solution from a hydrate

A solution with [Mg²⁺] = 0.200 M is to be made in a 250.0 mL volumetric flask from $\ce{MgSO4.7H2O}$ (246.48 g/mol). What mass of crystals is needed? If a student mistakenly used the molar mass of anhydrous $\ce{MgSO4}$ (120.37 g/mol) for the calculation but weighed out the hydrate, what would [Mg²⁺] be?

Step-by-step solution

Moles needed. 1 mol of crystal contains 1 mol of Mg²⁺:
$\begin{aligned} n &= MV = 0.200 \times 0.2500 \\ &= 0.0500\ \text{mol} \end{aligned}$
Mass of crystals. The water of crystallization counts in the molar mass:
$\begin{aligned} m &= 0.0500 \times 246.48 \\ &= \boxed{12.3\ \text{g}} \end{aligned}$
Procedure. Weigh the crystals and dissolve them in a little water in a beaker; transfer to the 250 mL volumetric flask, rinse the beaker 2–3 times and add the rinsings to the flask; fill to the mark and mix. Once dissolved, the water of crystallization is just part of the solvent and doesn’t affect filling to the mark.
The cost of the wrong molar mass. Using 120.37 g/mol, the student weighs only 0.0500 × 120.37 = 6.02 g of hydrate:
$\begin{aligned} n &= \frac{6.02}{246.48} = 0.0244\ \text{mol} \\ [\ce{Mg^2+}] &= \frac{0.0244}{0.2500} = \boxed{0.0977\ \text{M}} \end{aligned}$
That is only about half the target. When weighing a hydrate, always use the hydrate’s molar mass.

Practice: what is the concentration of this ion?

$\ce{Cl-}$ in 0.20 M NaCl
$\ce{Cl-}$ in 0.20 M $\ce{MgCl2}$
$\ce{Na+}$ in 0.15 M $\ce{Na2SO4}$
$\ce{SO4^2-}$ in 0.10 M $\ce{Na2SO4}$
$\ce{NO3-}$ in 0.10 M $\ce{Al(NO3)3}$
$\ce{Al^3+}$ in an $\ce{Al(NO3)3}$ solution with $[\ce{NO3-}]$ = 0.30 M
$\ce{I-}$ in 0.20 M $\ce{SrI2}$
$\ce{OH-}$ in 0.15 M $\ce{Ba(OH)2}$
$\ce{K+}$ after 0.40 M KCl is diluted to twice its volume
$\ce{Cl-}$ after equal volumes of 0.20 M NaCl and 0.20 M KCl are mixed
$\ce{Na+}$ after equal volumes of 0.20 M NaCl and 0.20 M KCl are mixed

Common mistakes

Mistake 1: “A 1.00 M solution is 1 mol of solute plus 1 L of water.” The 1 L is the volume of solution. Dissolve 1 mol of solute and add water in a volumetric flask up to the 1 L mark.

Mistake 2: “An ion’s concentration equals the solute’s concentration.” Multiply by the number of that ion in the formula. In 0.20 M $\ce{MgCl2}$ , $[\ce{Cl-}]$ is 0.40 M.

Mistake 3: “V₂ in M₁V₁ = M₂V₂ is the volume of water added.” V₂ is the total volume of the diluted solution. The water added is (approximately) V₂ − V₁.

Quick check

20.0 mL of 0.150 M $\ce{CaCl2}$ is mixed with 30.0 mL of 0.100 M NaCl. What is $[\ce{Cl-}]$ in the mixture?

Which is the correct way to prepare 1.00 L of 0.100 M NaCl (NaCl 58.44 g/mol)?

To measure out exactly 25.00 mL of a solution and transfer it to another container, which is the most suitable piece of glassware?

Which is the correct way to prepare 100.0 mL of 0.500 M solution from a 2.00 M stock solution?

All of these solutions are 0.10 M. Which conducts electricity best?

An $\ce{Al(NO3)3}$ solution has $[\ce{NO3-}]$ = 0.45 M. What is $[\ce{Al^3+}]$ ?

0.0350 mol of KCl is dissolved in water to make 250.0 mL of solution. How should its concentration be reported?

AP-style free response

(a) How many grams of $\ce{Na2CO3}$ (105.99 g/mol) are needed to prepare 250.0 mL of 0.100 M $\ce{Na2CO3}$ ?

(b) Briefly describe how to prepare this solution, naming the key piece of glassware.

(c) 25.0 mL of this solution is mixed with 25.0 mL of 0.100 M $\ce{NaHCO3}$ . Find $[\ce{Na+}]$ in the mixture.

(d) Which conducts better, 0.100 M $\ce{Na2CO3}$ or 0.100 M $\ce{NaHCO3}$ ? Justify your answer.

Model answer and scoring

(a) n = 0.100 × 0.2500 = 0.0250 mol, and mass = 0.0250 × 105.99 ≈ 2.65 g.

(b) Weigh 2.65 g of $\ce{Na2CO3}$ on an analytical balance, dissolve it in a little distilled water in a beaker, transfer it to a 250 mL volumetric flask (rinsing the beaker into the flask), add distilled water to the mark, then stopper and mix.

(c) n( $\ce{Na+}$ ) = 0.0250 × 0.100 × 2 + 0.0250 × 0.100 × 1 = 0.00500 + 0.00250 = 0.00750 mol; the total volume is 0.0500 L, so $[\ce{Na+}]$ = 0.150 M.

(d) $\ce{Na2CO3}$ . Each formula unit of $\ce{Na2CO3}$ releases 3 ions and each $\ce{NaHCO3}$ only 2, so at the same concentration the $\ce{Na2CO3}$ solution has the higher total ion concentration.

Scoring: (a) 2.65 g; (b) dissolve first, then bring to volume at the mark of a volumetric flask; (c) 0.150 M, counting 2 $\ce{Na+}$ in $\ce{Na2CO3}$ ; (d) justified by total ion concentration.

Summary

Molarity = moles of solute ÷ liters of solution; an ion’s concentration = the solute’s concentration × that ion’s number in the formula.
Mixing solutions that don’t react: add the moles of the same ion, then divide by the total volume.
Prepare solutions in a volumetric flask, dissolving first and then adding water to the mark; measure accurately with a volumetric pipet. On dilution, M₁V₁ = M₂V₂, where V₂ is the total volume of solution.
The higher the total ion concentration, the better a solution conducts: strong electrolytes > weak electrolytes > nonelectrolytes.

AP tip: To find ion concentrations, write the dissociation equation first so you see how many of each ion there are. When mixing solutions, add moles, never concentrations, and round to the right significant figures at the end. Lab questions often ask which step was done wrong or which glassware to use.

Solutions and Mixtures溶液与混合物

Learning goals

Before you start

A solution is a homogeneous mixture

Molarity and ion concentrations

Mixing two solutions

Preparing and diluting solutions

Electrolytes and conductivity

Example 1: [Cl⁻] after mixing

Example 2: Preparing a dilute solution from a concentrated one

Example 3: Preparing a solution from a hydrate

Common mistakes

Quick check

AP-style free response

Summary

Feedback / Report a problem