Prepare solutions from solid solutes and find their molarity, dilute a stock solution with M₁V₁ = M₂V₂, and determine an unknown concentration by acid–base titration — including diprotic acids. In each section, calculate the answer yourself first, then check it against the simulation.
Theory — Concentration, Dilution and Titration
Molar Concentration (Molarity)
Molarity (M) is the number of moles of solute per litre of solution. To prepare a solution from a solid, convert the mass to moles using the molar mass, then divide by the volume.
Molarity
moles = mass (g) ÷ molar mass (g/mol)
molarity (mol/L) = moles ÷ volume (L)
mass = moles × molar mass · moles = molarity × volume
Dilution
Adding solvent lowers the concentration but keeps the number of moles the same, so the moles before equal the moles after.
Dilution
M₁V₁ = M₂V₂
V₁ = M₂V₂ / M₁ (volume of stock needed)
Titration
In a titration, a solution of known concentration (the titrant) is added until it exactly reacts with the unknown — the equivalence point. For an acid–base titration, equate the moles of H⁺ and OH⁻. For a diprotic acid (such as H₂SO₄) each mole provides two moles of H⁺, so include the number of protons, n.
Titration at the Equivalence Point
Monoprotic: Mₐ Vₐ = M_b V_b
General: nₐ Mₐ Vₐ = n_b M_b V_b
(n = number of H⁺ for the acid, or OH⁻ for the base)
Diprotic acid (H₂SO₄): nₐ = 2
Molarity
mol of solute per litre. Convert mass→moles→divide by volume.
Dilution
M₁V₁ = M₂V₂; moles stay constant when solvent is added.
Titration
At equivalence, moles of acid H⁺ = moles of base OH⁻ (use n for polyprotic).
Compound
Molar mass (g/mol)
NaCl
58.44
KNO₃
101.10
CuSO₄
159.61
Glucose (C₆H₁₂O₆)
180.16
KMnO₄
158.03
Apparatus
The equipment a real titration experiment uses. In the simulation these are modelled for you, but the readings correspond to what each instrument would measure.
Burette
Delivers the standard solution to the endpoint.
Volumetric flask
Prepares solutions to an exact known volume.
Volumetric pipette
Transfers a precise aliquot of the unknown.
Erlenmeyer flask
Holds the sample being titrated.
Indicator
Signals the endpoint by a colour change.
Analytical balance
Weighs the primary standard to set the concentration.
Instructions — Running the Virtual Experiment
The simulation has three sections. In each one, calculate the answer yourself first, then use the simulation to check it, and record both values in your lab report with screenshots.
Part 1 — Make a Solution (Make a Solution tab)
1
Open Simulation → Make a Solution. Choose a solid solute, set the mass and the volume of water. Calculate the molarity yourself first (moles = mass ÷ molar mass; molarity = moles ÷ volume), then read the value the simulation shows and record both.
2
Do this for at least five different solid chemicals. Also work out the mass needed to make a chosen molarity, and compare.
Part 2 — Dilution (Dilution tab)
1
Open Dilution. Choose what to find. To get the diluted concentration, set M₁, V₁, and V₂ and calculate M₂ = M₁V₁/V₂. To get the stock volume needed, set M₁, the target M₂, and V₂ and calculate V₁ = M₂V₂/M₁. Work it out yourself first, then check and record both.
2
Do at least two dilutions — for example, one of each type.
Part 3 — Titration (Titration tab)
1
Open Titration. Choose the acid (monoprotic or diprotic) and set the known base concentration and the volumes. Calculate the unknown acid concentration yourself first using nₐMₐVₐ = n_bM_bV_b, then run the titration to the equivalence point and record both.
2
Do at least one monoprotic and one diprotic titration, and note how the diprotic acid needs twice as much base for the same concentration.
Simulation — Concentration, Dilution and Titration
Solutions Virtual LabMake · Dilute · Titrate
0.34 mol/L NaCl
Prepare a solution
Simulation value
Molar mass58.44 g/mol
Moles—
Concentration—
Calculate first, then show the simulation value.
stock
→
diluted
Dilute a stock solution
Simulation value
M₁V₁—
Diluted conc. M₂—
Calculate the answer first, then check.
Add base to reach the equivalence point
Acid–base titration
Result at equivalence
n (acid)1
moles base—
moles acid H⁺—
[Acid] unknown—
nₐMₐVₐ = n_bM_bV_b at the equivalence point.
Team Questions
Question 1. What is the molarity of a solution made by dissolving 20.0 g of NaCl (58.44 g/mol) in 1.00 L of water? (Type to 3 decimals)
Question 2. How many grams of glucose (180.16 g/mol) are needed to make 1.00 L of 0.100 mol/L solution? (Type in g, to 1 decimal)
Question 3. You dilute 50.0 mL of 1.00 mol/L solution to a final volume of 250 mL. What is the new concentration? (M₁V₁ = M₂V₂; type to 3 decimals)
Question 4. 25.0 mL of 0.100 mol/L NaOH exactly neutralises 20.0 mL of HCl. What is [HCl]? (Type to 3 decimals)
Question 5. 40.0 mL of 0.100 mol/L NaOH neutralises 20.0 mL of H₂SO₄ (diprotic). What is [H₂SO₄]? (Use 2·Mₐ·Vₐ = M_b·V_b; type to 3 decimals)
Question 6 — Challenge. Why does a diprotic acid need twice as much base as a monoprotic acid of the same concentration and volume? (Answer in a phrase)
Question 7. How many moles of solute are in 250 mL of 0.40 mol/L solution? (moles = M × V in litres; type to 2 decimals)
Example Lab Report
Sample report demonstrating the expected format. Include your data table of solutions, the calculated vs simulated comparison, and worked titration calculations, with labelled screenshots.
To prepare solutions of known molarity from solid solutes, to dilute a stock solution using M₁V₁ = M₂V₂, and to determine an unknown acid concentration by titration, including a diprotic acid.
Theory
Molarity is moles of solute per litre. Diluting keeps the moles constant. At the equivalence point of a titration the moles of H⁺ equal the moles of OH⁻, with a factor n for polyprotic acids.
The calculated molarities matched the simulation values. Compounds with larger molar masses gave lower molarity for the same 20 g, because fewer moles are present. The diprotic acid required twice the volume of base compared with a monoprotic acid of the same concentration, since each H₂SO₄ supplies two protons.
Conclusion
Molarity links mass, molar mass, and volume; dilution conserves moles; and titration determines an unknown concentration at the equivalence point, with the number of protons taken into account for polyprotic acids.
Practice Questions
Show all work. Use M = (mass/molar mass)/V, M₁V₁ = M₂V₂, and nₐMₐVₐ = n_bM_bV_b.
Question 1
Calculate the molarity when 20.0 g of KNO₃ (101.10 g/mol) is dissolved in 1.00 L of water.
Hint: moles = 20.0/101.10 = 0.198; M = 0.198 mol/L.
Question 2
What mass of CuSO₄ (159.61 g/mol) is needed for 0.500 L of a 0.200 mol/L solution?
Hint: moles = 0.200 × 0.500 = 0.100; mass = 0.100 × 159.61 = 15.96 g.
Question 3
How would you prepare 500 mL of 0.10 mol/L solution from a 2.0 mol/L stock?
Hint: V₁ = M₂V₂/M₁ = (0.10 × 500)/2.0 = 25 mL stock, then top up to 500 mL.
Question 4
A 20.0 mL sample of HCl is titrated with 0.150 mol/L NaOH; 18.0 mL is needed. Find [HCl].
Hint: [HCl] = (0.150 × 18.0)/20.0 = 0.135 mol/L.
Question 5
A 25.0 mL sample of H₂SO₄ (diprotic) needs 30.0 mL of 0.200 mol/L NaOH. Find [H₂SO₄].
