E. Solutions and Solubility · Solutions and Solubility

Solution Stoichiometry

E2.6 E2.2 E2.1 A1.5 A1.8 A1.13About 75 minutes

Learning goals

I can use n = cV with mole ratios from balanced equations to find volumes, concentrations and masses in solution reactions.
I can calculate the mass of precipitate formed, and use a precipitate's mass to find an unknown concentration (gravimetric analysis).
I can identify the limiting reagent when two solutions are mixed and calculate theoretical and percentage yield.
I can identify sources of error in a precipitation experiment.

Words to know

solution stoichiometry
Calculating quantities of reactants and products in reactions that take place in solution, using n = cV.

mole ratio
The ratio of coefficients in a balanced equation, used to convert moles of one substance to moles of another.

limiting reagent
The reactant that is used up first and so determines how much product forms.

excess reagent
A reactant that is left over after the reaction is complete.

gravimetric analysis
Finding the amount of a substance by forming a precipitate, then filtering, drying and weighing it.

percentage yield
(actual yield ÷ theoretical yield) × 100%.

How much do you need?

A water treatment operator must decide how many litres of iron(III) chloride solution to add to remove the phosphate from a day's sewage. A pharmacist checks how much antacid neutralizes a given amount of stomach acid. A lab technician weighs a precipitate to find the sulfate in a lake sample. All three use solution stoichiometry: the stoichiometry you learned in the Quantities in Chemical Reactions unit, with n = cV as the way in and out of moles.

Remember from the Quantities in Chemical Reactions unit

Stoichiometry uses the coefficients of a balanced equation as a mole ratio. The steps are: find moles of the given substance → multiply by the mole ratio → convert moles of the required substance into the unit asked for. You also learned to find the limiting reagent and percentage yield. In solutions, the only new step is n = cV.

A road map for solution stoichiometry. Whatever you are given, convert it to moles, use the mole ratio, then convert to what is asked.

Worked example

Example 1: Volume needed to neutralize

What volume of 0.250 mol/L NaOH(aq) is needed to neutralize 25.0 mL of 0.100 mol/L H₂SO₄(aq)?

Balanced equation: H₂SO₄(aq) + 2 NaOH(aq) → Na₂SO₄(aq) + 2 H₂O(l).
Given: n(H₂SO₄) = cV = 0.100 mol/L × 0.0250 L = 0.00250 mol.
Mole ratio: n(NaOH) = 0.00250 mol × (2 mol NaOH ÷ 1 mol H₂SO₄) = 0.00500 mol.
Required volume: V = n ÷ c = 0.00500 mol ÷ 0.250 mol/L = 0.0200 L = 20.0 mL.

20.0 mL of NaOH(aq)

Worked example

Example 2: Mass of precipitate

Excess KI(aq) is added to 50.0 mL of 0.200 mol/L Pb(NO₃)₂(aq). What mass of PbI₂ precipitate forms? (M(PbI₂) = 461.0 g/mol)

Equation: Pb(NO₃)₂(aq) + 2 KI(aq) → PbI₂(s) + 2 KNO₃(aq). (Net ionic: Pb²⁺ + 2 I⁻ → PbI₂(s).)
n(Pb(NO₃)₂) = 0.200 × 0.0500 = 0.0100 mol. KI is in excess, so lead(II) nitrate is limiting.
Mole ratio 1 : 1, so n(PbI₂) = 0.0100 mol.
m = nM = 0.0100 mol × 461.0 g/mol = 4.61 g.

4.61 g of PbI₂

Gravimetric analysis: finding an unknown concentration

The reverse of Example 2 is a powerful lab method. To find how much of an ion is in a sample:

Add an excess of a reagent that precipitates that ion completely.
Filter to collect the precipitate (pre-weigh the filter paper).
Rinse the precipitate with a little distilled water to remove dissolved spectator ions.
Dry it completely and weigh it; dry and weigh again until the mass stops changing.
Use stoichiometry backward to find the moles, then the concentration, of the original ion.

The four steps of gravimetric analysis. The final mass (here 0.375 g) is the measured value used in Example 3.

Worked example

Example 3: Gravimetric analysis

Excess Na₂CO₃(aq) is added to 25.0 mL of a CaCl₂ solution. The dried CaCO₃ precipitate has a mass of 0.375 g. Find [CaCl₂]. (M(CaCO₃) = 100.09 g/mol)

Equation: CaCl₂(aq) + Na₂CO₃(aq) → CaCO₃(s) + 2 NaCl(aq).
n(CaCO₃) = m ÷ M = 0.375 g ÷ 100.09 g/mol = 0.003747 mol.
Mole ratio 1 : 1, so n(CaCl₂) = 0.003747 mol.
c = n ÷ V = 0.003747 mol ÷ 0.0250 L = 0.150 mol/L.

[CaCl₂] = 0.150 mol/L

Worked example

Example 4: Limiting reagent and percentage yield

30.0 mL of 0.400 mol/L AgNO₃(aq) is mixed with 25.0 mL of 0.300 mol/L NaCl(aq). (a) Which is limiting? (b) Find the theoretical mass of AgCl (143.32 g/mol). (c) A student collects 1.02 g of dry AgCl. Find the percentage yield.

Equation: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq) (1 : 1).
n(AgNO₃) = 0.400 × 0.0300 = 0.0120 mol. n(NaCl) = 0.300 × 0.0250 = 0.00750 mol.
(a) They react 1 : 1, and there is less NaCl, so NaCl is limiting; AgNO₃ is in excess.
(b) n(AgCl) = 0.00750 mol; m = 0.00750 × 143.32 = 1.07 g.
(c) % yield = 1.02 ÷ 1.075 × 100% = 94.9%.

NaCl is limiting; theoretical yield 1.07 g; percentage yield 94.9%.

Sources of error in precipitation experiments (A1.8)

Precipitate passes through the filter (particles too fine, torn paper): mass too low.
Precipitate not completely dry (water still in it): mass too high, sometimes giving a yield over 100%.
Not rinsed: dissolved spectator ions dry onto the precipitate: mass too high.
Some precipitate left in the beaker: mass too low.
Not enough reagent (the 'excess' was not really excess): mass too low.

A percentage yield above 100% is a clue that the product is not pure or not dry.

Just for fun

The word stoichiometry comes from Greek words meaning "element" and "measure." Saying it correctly (stoy-key-OM-uh-tree) is worth at least one bonus mark of confidence.

Hands-on activity

Kitchen stoichiometry: how much baking soda does vinegar need?

You need: white vinegar (5% acetic acid), baking soda, a kitchen scale (or ¼-teaspoon measure), a tall glass, a spoon, a measuring cup, and a sink.

Predict: 50 mL of 5% m/v vinegar contains 2.5 g of acetic acid (60.06 g/mol), or 0.042 mol. The reaction is NaHCO₃ + CH₃COOH → NaCH₃COO + H₂O + CO₂ (1 : 1). Calculate the mass of baking soda (84.01 g/mol) that exactly reacts: 0.042 × 84.01 ≈ 3.5 g (about ¾ teaspoon).
Put 50 mL of vinegar in the glass, in the sink. Add baking soda in small portions (¼ teaspoon), stirring, and wait for fizzing to stop each time.
Keep track of the total added. Stop when a new portion causes no more fizzing and some white powder stays undissolved.
Compare your total with the prediction. Which reactant was limiting at each stage?

What to look for: fizzing stops near 3.5 g (roughly ¾ to 1 teaspoon). Before that point, baking soda is the limiting reagent; after it, the acetic acid is used up and baking soda is in excess. Give two reasons your result might differ (vinegar strength, scale precision, CO₂ staying dissolved). Everything can be rinsed down the drain.

Your turn

What volume of 0.100 mol/L NaOH is needed to neutralize 20.0 mL of 0.150 mol/L H₂SO₄?

Excess BaCl₂ is added to 40.0 mL of 0.250 mol/L Na₂SO₄. What mass of BaSO₄ (233.40 g/mol) forms?

Excess NaOH is added to 50.0 mL of 0.200 mol/L CuSO₄. Mass of Cu(OH)₂ (97.57 g/mol)?

What volume of 0.500 mol/L AgNO₃ precipitates all the Cl⁻ in 25.0 mL of 0.200 mol/L CaCl₂?

25.0 mL of HCl(aq) is exactly neutralized by 18.4 mL of 0.125 mol/L NaOH. Find [HCl].

Why can't you go straight from volume of one solution to volume of the other without moles?

More practice: warm-up

How many moles are in 15.0 mL of 0.200 mol/L HCl?

In 2 NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O, what is the mole ratio NaOH : H₂SO₄?

What is gravimetric analysis?

Why is a reagent added in excess in gravimetric analysis?

Write the formula for percentage yield.

More practice: on your own

25.0 mL of 0.120 mol/L Ba(OH)₂ is neutralized by 95.0 mL of HCl. Find [HCl].

Excess AgNO₃ is added to 50.0 mL of NaBr(aq), giving 1.25 g of AgBr (187.77 g/mol). Find [NaBr].

20.0 mL of 0.500 mol/L BaCl₂ is mixed with 30.0 mL of 0.200 mol/L Na₂SO₄. Find the limiting reagent and the theoretical mass of BaSO₄.

In the last question a student collects 1.31 g. Find the percentage yield and suggest one reason it is below 100%.

A sewage plant must remove phosphate from 1.00 × 10³ L of water containing 5.0 × 10⁻⁴ mol/L PO₄³⁻. What mass of FeCl₃ (162.20 g/mol) is needed? (Fe³⁺ + PO₄³⁻ → FePO₄)

Challenge

An antacid tablet contains 0.500 g of Mg(OH)₂ (58.33 g/mol). What volume of 0.10 mol/L HCl (similar to stomach acid) can it neutralize?

50.0 mL of 0.100 mol/L AgNO₃ is mixed with 50.0 mL of 0.0800 mol/L NaCl. Find the mass of AgCl formed and the concentrations of all ions left in solution.

A student reports a percentage yield of 108% for a precipitation. Give two likely causes and explain how to fix each.

Check your understanding

Quiz

How many moles of NaOH are in 25.0 mL of 0.200 mol/L NaOH?

What volume of 0.200 mol/L NaOH neutralizes 10.0 mL of 0.100 mol/L HCl?

H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O. 0.010 mol of H₂SO₄ needs

0.010 mol AgNO₃ is mixed with 0.020 mol NaCl. The limiting reagent is

A precipitate that is weighed while still wet gives a calculated concentration that is

Theoretical yield 2.00 g, actual yield 1.80 g. Percentage yield:

For parents and teachers

Big idea (E2.6): solution stoichiometry is ordinary stoichiometry with n = cV as the entry and exit step.

How to help: Ask your teen to write the road map (given → moles → mole ratio → moles → answer) before touching the calculator, and to label every number with its unit and formula.

Common difficulties:

Skipping the mole ratio (especially 2 : 1 ratios with H₂SO₄ or CaCl₂).
Using mL in n = cV.
Forgetting that the total volume changes when two solutions are mixed (needed for leftover ion concentrations).
Choosing the limiting reagent by comparing volumes or masses instead of moles adjusted by the ratio.

This lesson builds directly on stoichiometry, limiting reagents and percentage yield from the Quantities in Chemical Reactions unit. The kitchen activity is safe; do it over a sink.