Volumetric Calculations involves the use of both gravimetric analysis and stoichiometry.
Example #1: When dealing with volumetric calculations we utilize a stoichiometric chart.
Example #2: Iron (III) can be oxidized by an acidic K2Cr2O7 solution according to the net ionic equation below. How many microliters of a 0.250 M FeCl2 are needed to completely react with 9.12 g of a compound containing 41.5% weight K2Cr2O7?
Cr2O72- + 6 Fe2+ + 14 H+ → 2 Cr3+ + 6 Fe3+ + 7 H2O
Example #3: Magnesium reacts with HCl according to the reaction below. How many grams of 5.310% by weight of aqueous magnesium are required to provide a 25% excess to react with 75.0 mL of 0.0550 M HCl.
Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g)
Example #4: The amount of iron within an ore sample was determined by an oxidation-reduction titration using potassium permanganate, KMnO4, as the titrant. A 0.5600 g sample of the ore was placed into acid and the newly freed Fe3+ was then reduced to Fe2+. The titration of this solution required 39.82 mL of 0.0315 M KMnO4 to reach the end-point. Determine the mass percent of Fe2O3 in the sample.
MnO4– + 5 Fe2+ + 8 H+ ⇌ Mn2+ + 5 Fe3+ + 4 H2O
Example #5: A 0.4317 g sample of CaCO3 (MW: 100.09 g/mol) is added to flask that also contained 12.50 mL of 1.530 M HBr.
CaCO3 (aq) + 2 HBr (aq) → CaBr2 (aq) + H2O (l) + CO2 (g)
Additional water is then added to create a 250.0 mL of Solution A. Next 20.00 mL aliquot of solution A is taken and titrated with 0.0980 M NaOH. How many milliliters of NaOH were used?
NaOH (aq) + HBr (aq) → H2O (l) + NaBr (aq)
Example #6: A 9.2476 g sample of M(OH)2 was mixed with 15.00 mL of 1.530 M HI and diluted to a final 125.0 mL of solution.
M(OH)2 (aq) + 2 HI (aq) → 2 H2O (l) + MCl2 (aq)
A 12.00 mL aliquot of this diluted solution was taken and titrated with 18.23 mL of 0.0695 M NaOH.
NaOH (aq) + HI (aq) → H2O (l) + NaCl (aq)
What is the identity of the metal representing M?
Example #7: A 6.2034 g sample of Cu(OH)n was mixed with 25.00 mL of 2.250 M Hl and diluted to a final 250.0 mL of solution.
Cu(OH)n (s) + n HI (aq) → CuIn (aq) + n H2O (l)
A 50.00 mL aliquot of this solution was taken and titrated with 16.25 mL of 0.1250 M KOH. What is the value of n?
KOH (aq) + HI (aq) → H2O (l) + NaI (aq)
Practice: A 1.000 g sample of Na2CO3 (MW: 105.99 g/mol) is dissolved in enough water to make 200.0 mL of solution. A 25.00 mL aliquot required 32.18 mL of HCl to completely neutralize it. What is the molar concentration of HCl?
Na2CO3 (aq) + 2 HCl (aq) → 2 KCl (aq) + H2O(l) + CO2 (g)