1 question, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 1
A 10.00 mL sample of H₂SO₄(aq) of unknown concentration is titrated with 0.2000 M NaOH(aq) to a phenolphthalein endpoint. The reaction is H₂SO₄(aq) + 2 NaOH(aq) → Na₂SO₄(aq) + 2 H₂O(l). The initial buret reading is 10.20 mL and the final buret reading is 25.20 mL. What is the molarity of the H₂SO₄(aq)?
Answer and reasoning
A0.3000 M A student who thinks the moles of titrant and analyte are equal at the equivalence point picks this, taking 3.000 × 10⁻³ mol H₂SO₄. The balanced equation shows that each H₂SO₄ reacts with two NaOH, so only 1.500 × 10⁻³ mol H₂SO₄ was present.
B0.1500 MCorrect The NaOH delivered is 25.20 − 10.20 = 15.00 mL, or 3.000 × 10⁻³ mol. The equation needs 2 mol NaOH per mole of H₂SO₄, so n(H₂SO₄) = 1.500 × 10⁻³ mol in 0.01000 L: 0.1500 M.
C0.2520 M A student who takes the final buret reading as the volume delivered picks this, using 25.20 mL. The buret started at 10.20 mL, so only 15.00 mL of NaOH was delivered.
D0.2000 M A student who thinks the analyte and titrant have equal concentrations at the equivalence point picks this. The equivalence point is defined by moles in the stoichiometric ratio; the H₂SO₄ concentration must be calculated from the moles of NaOH used.
Working Volume of NaOH delivered = 25.20 − 10.20 = 15.00 mL. n(NaOH) = 0.01500 L × 0.2000 M = 3.000 × 10⁻³ mol. n(H₂SO₄) = ½ × 3.000 × 10⁻³ = 1.500 × 10⁻³ mol. M = 1.500 × 10⁻³ mol ÷ 0.01000 L = 0.1500 M.
In preparation: 0 of 1 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
4.6.A.1 Titration Fix
Titration
A procedure that determines the amount of a substance (the analyte) in a solution by adding measured volumes of a solution of known concentration (the titrant) that reacts specifically and quantitatively with it, until the analyte is used up.
Titrant
The solution of known concentration delivered from a buret during a titration; the moles of titrant used are found from its concentration and the volume delivered (final buret reading minus initial reading).
Analyte
The substance whose amount (or concentration) is being determined in a titration. Adding water to the analyte sample changes its concentration but not the moles of analyte present.
Equivalence point
The point in a titration at which the analyte has been totally consumed by the reacting species in the titrant: the moles of titrant added and the moles of analyte are in the mole ratio of the balanced equation, which need not be 1 : 1.
Endpoint
The observable event, such as a color change of an indicator or of the titrant itself, that signals that the equivalence point has been reached. In a well-designed titration the endpoint is very close to the equivalence point; the experimenter stops at the first lasting change.
Indicator
A substance added in a very small amount whose color changes at or very near the equivalence point; it is not a significant reactant, and its own amount is not used in the calculation.
Following a titration by a measured property
Besides a color change, the equivalence point can be located from a property measured during the titration, such as electrical conductivity: the property changes trend sharply when the analyte is used up and excess titrant begins to accumulate.
Students often think At the equivalence point the moles of titrant added equal the moles of analyte, whatever the balanced equation. In fact Only when the balanced equation has a 1 : 1 mole ratio. For H₂SO₄ + 2 NaOH, the equivalence point comes when the moles of NaOH added are twice the moles of H₂SO₄.
Students often think The final buret reading is the volume of titrant delivered. In fact Only if the initial reading was 0.00 mL. The volume delivered is the final reading minus the initial reading.
5 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 5
A 50.0 mL sample of Ba(OH)₂(aq) is titrated with 0.100 M HCl(aq), and the electrical conductivity of the solution in the flask is measured after each addition. The reaction is Ba(OH)₂(aq) + 2 HCl(aq) → BaCl₂(aq) + 2 H₂O(l). Based on the graph, how many millimoles (1 mmol = 10⁻³ mol) of Ba(OH)₂ were in the sample?
Answer and reasoning
A2.00 mmol A student who thinks the moles of titrant and analyte are equal at the equivalence point picks this. The equation shows 2 mol HCl for each mole of Ba(OH)₂, so 2.00 mmol HCl reacts with 1.00 mmol Ba(OH)₂.
B1.75 mmol A student who thinks all of the titrant added reacts with the analyte picks this, using the last volume, 35.0 mL. After 20.0 mL the Ba(OH)₂ is used up; the HCl added after that does not react, which is why the conductivity rises.
C1.00 mmolCorrect The equivalence point is where the conductivity stops falling and starts rising, at 20.0 mL: up to there each portion of HCl removes OH⁻; afterward the added HCl has nothing to react with. n(HCl) = 20.0 mL × 0.100 M = 2.00 mmol, and the equation needs 2 HCl per Ba(OH)₂, so n(Ba(OH)₂) = 1.00 mmol.
D2.50 mmol A student who thinks the equivalence point comes when the titrant volume equals the 50.0 mL sample volume picks this. The equivalence point is where the analyte is used up, shown by the change in trend at 20.0 mL.
Working The conductivity changes from falling to rising at 20.0 mL: this is the equivalence point, where all the OH⁻ has been consumed; after it, added HCl simply accumulates. n(HCl) = 0.0200 L × 0.100 M = 2.00 × 10⁻³ mol; n(Ba(OH)₂) = ½ × 2.00 × 10⁻³ = 1.00 × 10⁻³ mol = 1.00 mmol.
A sample of HCl(aq) is titrated with NaOH(aq), using phenolphthalein as the indicator. Which numbered diagram best represents the solute particles in a small volume of the solution in the flask at the equivalence point? Water molecules are not shown.
Answer and reasoning
ADiagram 1Correct At the equivalence point the HCl has been totally consumed by the OH⁻ added, forming water, with no OH⁻ left over. The solute particles left are Na⁺ from the titrant and Cl⁻ from the analyte, equal in number and separate from each other.
BDiagram 2 A student who thinks equal amounts of acid and base remain together at the equivalence point picks this. H₃O⁺ and OH⁻ react with each other to form water, so they cannot both remain in amounts comparable to the Na⁺ and Cl⁻ ions.
CDiagram 3 A student who thinks a dissolved ionic product exists as molecules picks this. NaCl is ionic: in solution the Na⁺ and Cl⁻ ions are separate, each surrounded by water molecules.
DDiagram 4 A student who treats the endpoint color, which needs a slight excess of base, as defining the equivalence point picks this. The diagram shows OH⁻ and extra Na⁺ from titrant added beyond the equivalence point; at the equivalence point the moles of OH⁻ added equal the moles of HCl, so none is left over.
Working No calculation. H₃O⁺(aq) + OH⁻(aq) → 2 H₂O(l). At the equivalence point all of the HCl has been consumed by exactly the stoichiometric amount of OH⁻: no significant H₃O⁺ or OH⁻ remains, and the Na⁺ and Cl⁻ ions remain in equal numbers as separate ions.
A nearly colorless solution of Fe²⁺(aq) is titrated with purple KMnO₄(aq). As it reacts, MnO₄⁻ is converted to Mn²⁺, which is nearly colorless. No purple or pink color remains in the flask until the endpoint, when a faint pink color persists. Which explanation of these observations is correct?
Answer and reasoning
AMnO₄⁻ does not react with Fe²⁺ until enough titrant has been added to reach the endpoint A student who thinks the titration reaction happens only at the endpoint picks this. If the MnO₄⁻ did not react, its purple color would appear with the first drop; the solution stays colorless because each portion reacts as it is added.
BThe color persists once the volume of KMnO₄ added equals the Fe²⁺ sample volume A student who thinks the equivalence point comes when equal volumes have been combined picks this. The color persists when the moles of MnO₄⁻ added reach one-fifth of the moles of Fe²⁺, whatever the volumes.
CThe color persists once the concentration of MnO₄⁻ added equals that of the Fe²⁺ A student who thinks the analyte and titrant concentrations are equal at the equivalence point picks this. The endpoint depends on amounts in the 1 : 5 mole ratio, not on any equality of concentrations.
DEach MnO₄⁻ added reacts with Fe²⁺ until all Fe²⁺ is consumed, and then MnO₄⁻ remainsCorrect MnO₄⁻ reacts with Fe²⁺ as soon as it is added, and its product, Mn²⁺, is nearly colorless. Once the Fe²⁺ is totally consumed (the equivalence point), the next MnO₄⁻ added does not react, so its color persists: this observable change is the endpoint.
Working No calculation. MnO₄⁻(aq) + 5 Fe²⁺(aq) + 8 H⁺(aq) → Mn²⁺(aq) + 5 Fe³⁺(aq) + 4 H₂O(l). While Fe²⁺ remains, each portion of MnO₄⁻ added is consumed at once, so its purple color does not persist. When all the Fe²⁺ has been consumed (the equivalence point), the next drop of MnO₄⁻ has nothing to react with and its color shows: the endpoint.
A student titrates a sample of HCl(aq) with standardized NaOH(aq), using phenolphthalein as the indicator. Instead of stopping at the first faint pink color that persists, the student keeps adding NaOH until the solution is dark pink and records that buret reading. How does this error affect the calculated number of moles of HCl in the sample?
Answer and reasoning
AThey are not affected, because any pink color marks the equivalence point of the titration A student who thinks the endpoint is the equivalence point whatever color is seen picks this. The equivalence point is defined by amounts; the first lasting faint pink signals it, and NaOH added beyond that point is excess.
BThey are too high, because more NaOH was recorded than was needed to react with the HClCorrect The first lasting faint pink signals that the HCl has just been consumed. The extra NaOH added to reach dark pink did not react with HCl, but it is included in the recorded volume, so the moles of NaOH, and the moles of HCl calculated from them, are too high.
CThey are too low, because the extra NaOH solution dilutes the HCl that is in the flask A student who thinks diluting the analyte reduces the amount of analyte picks this. Adding liquid changes the concentration in the flask, not the moles of HCl; the error is in the recorded NaOH volume, which is too large.
DThey are not affected, because the moles of HCl in the flask were fixed beforehand A student who thinks a fixed actual amount means the calculated amount cannot be in error picks this. The actual moles of HCl are fixed, but the calculated value comes from the recorded NaOH volume, which includes excess NaOH.
Working No calculation. The first lasting faint pink marks the equivalence point. NaOH added after that does not react with HCl, but it is included in the recorded volume, so moles of NaOH, and therefore the calculated moles of HCl, are too high.
A sample of I₂(aq) is titrated with Na₂S₂O₃(aq) of known concentration. The titration reaction, which goes to completion, is I₂(aq) + 2 S₂O₃²⁻(aq) → 2 I⁻(aq) + S₄O₆²⁻(aq). Which statement must be true at the equivalence point of this titration?
Answer and reasoning
AThe moles of S₂O₃²⁻ added are twice the moles of I₂ in the sampleCorrect The equivalence point is reached when the I₂ has been totally consumed by the S₂O₃²⁻ added. The equation shows that each mole of I₂ reacts with 2 mol S₂O₃²⁻, so the moles of S₂O₃²⁻ added must be twice the moles of I₂ that were in the sample.
BThe moles of S₂O₃²⁻ added are equal to the moles of I₂ in the sample A student who thinks the moles of titrant and analyte are equal at the equivalence point picks this. With equal moles, only half of the I₂ would have reacted, because each I₂ needs two S₂O₃²⁻ ions.
CThe volume of Na₂S₂O₃(aq) added is equal to the volume of the I₂(aq) sample A student who thinks the equivalence point comes when equal volumes have been combined picks this. The volume of titrant needed depends on the moles of I₂ and on the concentration of the titrant; it matches the sample volume only by coincidence.
DThe volume of Na₂S₂O₃(aq) added is twice the volume of the I₂(aq) sample A student who uses the coefficients as a ratio of solution volumes picks this. The 1 : 2 ratio applies to moles; the volumes are in that ratio only if the two solutions happen to have the same molarity.
Working At the equivalence point the analyte, I₂, is totally consumed by the S₂O₃²⁻ added, with no S₂O₃²⁻ left over. The equation requires 2 mol S₂O₃²⁻ per mole of I₂, so n(S₂O₃²⁻ added) = 2 × n(I₂). In terms of the solutions, M(S₂O₃²⁻) × V(titrant) = 2 × M(I₂) × V(sample); no fixed relationship between the two volumes follows unless the concentrations are known.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
