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4.5 Quantitative Chemical Analysis
By the end of this section, you will be able to:
- Describe the fundamental aspects of titrations and gravimetric analysis.
- Perform stoichiometric calculations using typical titration and gravimetric data. In the 18th century, the strength (actually the concentration) of vinegar samples was determined by noting the amount of potassium carbonate, K 2 CO 3 , which had to be added, a little at a time, before bubbling ceased. The greater the weight of potassium carbonate added to reach the point where the bubbling ended, the more concentrated the vinegar. We now know that the effervescence that occurred during this process was due to reaction with acetic acid, CH 3 CO 2 H, the compound primarily responsible for the odor and taste of vinegar. Acetic acid reacts with potassium carbonate according to the following equation:
2CH 3 CO 2 H( aq ) + K 2 CO 3 ( s ) ⟶ KCH 3 CO 3 ( aq ) + CO 2 ( g ) + H 2 O( l )
The bubbling was due to the production of CO 2.
The test of vinegar with potassium carbonate is one type of quantitative analysis —the determination of the amount or concentration of a substance in a sample. In the analysis of vinegar, the concentration of the solute (acetic acid) was determined from the amount of reactant that combined with the solute present in a known volume of the solution. In other types of chemical analyses, the amount of a substance present in a sample is determined by measuring the amount of product that results.
Titration
The described approach to measuring vinegar strength was an early version of the analytical technique known as titration analysis. A typical titration analysis involves the use of a buret ( Figure 4.15 ) to make incremental additions of a solution containing a known concentration of some substance (the titrant ) to a sample solution containing the substance whose concentration is to be measured (the analyte ). The titrant and analyte undergo a chemical reaction of known stoichiometry, and so measuring the volume of titrant solution required for complete reaction with the analyte (the equivalence point of the titration) allows calculation of the analyte concentration. The equivalence point of a titration may be detected visually if a distinct change in the appearance of the sample solution accompanies the completion of the reaction. The halt of bubble formation in the classic vinegar analysis is one such example, though, more commonly, special dyes called indicators are added to the sample solutions to impart a change in color at or very near the equivalence point of the titration. Equivalence points may also be detected by measuring some solution property that changes in a predictable way during the course of the titration. Regardless of the approach taken to detect a titration’s equivalence point, the volume of titrant actually measured is called the end point. Properly designed titration methods typically ensure that the difference between the equivalence and end points is negligible. Though any type of chemical reaction may serve as the basis for a titration analysis, the three described in this chapter (precipitation, acid-base, and redox) are most common. Additional details regarding titration analysis are provided in the chapter on acid-base equilibria.
Figure 4.15 (a) A student fills a buret in preparation for a titration analysis. (b) A typical buret permits volume measurements to the nearest 0.01 mL. (credit a: modification of work by Mark Blaser and Matt Evans; credit b: modification of work by Mark Blaser and Matt Evans)
Example 4.
Titration Analysis
The end point in a titration of a 50.00-mL sample of aqueous HCl was reached by addition of 35.23 mL of 0.250 M NaOH titrant. The titration reaction is: HCl( aq ) + NaOH( aq ) ⟶ NaCl( aq ) + H 2 O( l )
What is the molarity of the HCl? Solution As for all reaction stoichiometry calculations, the key issue is the relation between the molar amounts of the chemical species of interest as depicted in the balanced chemical equation. The approach outlined in previous modules of this chapter is followed, with additional considerations required, since the amounts of reactants are provided and requested are expressed as solution concentrations. For this exercise, the calculation will follow the following outlined steps:
The required change of state in a gravimetric analysis may be achieved by various physical and chemical processes. For example, the moisture (water) content of a sample is routinely determined by measuring the mass of a sample before and after it is subjected to a controlled heating process that evaporates the water. Also common are gravimetric techniques in which the analyte is subjected to a precipitation reaction of the sort described earlier in this chapter. The precipitate is typically isolated from the reaction mixture by filtration, carefully dried, and then weighed ( Figure 4.16 ). The mass of the precipitate may then be used, along with relevant stoichiometric relationships, to calculate analyte concentration.
Figure 4.16 Precipitate may be removed from a reaction mixture by filtration.
Example 4.
Gravimetric Analysis
A 0.4550-g solid mixture containing CaSO 4 is dissolved in water and treated with an excess of Ba(NO 3 ) 2 , resulting in the precipitation of 0.6168 g of BaSO 4. CaSO 4 ( aq ) + Ba(NO 3 ) 2 ( aq ) ⟶ BaSO 4 ( s ) + Ca(NO 3 ) 2 ( aq )
What is the concentration (percent) of CaSO 4 in the mixture?
Solution
The plan for this calculation is similar to others used in stoichiometric calculations, the central step being the connection between the moles of BaSO 4 and CaSO 4 through their stoichiometric factor. Once the mass of CaSO 4 is computed, it may be used along with the mass of the sample mixture to calculate the requested percentage concentration.
The mass of CaSO 4 that would yield the provided precipitate mass is
0.6168 g BaSO 4 ×
1 mol BaSO 4 233.43 g BaSO 4
×
1 mol CaSO 4 1 mol BaSO 4
× 136.14 g CaSO^4 1 mol CaSO 4
= 0.3597 g CaSO 4
The concentration of CaSO 4 in the sample mixture is then calculated to be
percent CaSO 4 = mass CaSO mass sample^4 × 100 %
0.3597 g 0.4550 g × 100 %^ = 79.05 %
Check Your Learning What is the percent of chloride ion in a sample if 1.1324 g of the sample produces 1.0881 g of AgCl when treated with excess Ag+? Ag+( aq ) + Cl−( aq ) ⟶ AgCl( s ) Answer: 23.76%
The elemental composition of hydrocarbons and related compounds may be determined via a gravimetric method known as combustion analysis. In a combustion analysis, a weighed sample of the compound is heated to a high temperature under a stream of oxygen gas, resulting in its complete combustion to yield gaseous products of known identities. The complete combustion of hydrocarbons, for example, will yield carbon dioxide and water as the only products. The gaseous combustion products are swept through separate, preweighed collection devices containing compounds that selectively absorb each product ( Figure 4.17 ). The mass increase of each device corresponds to the mass of the absorbed product and may be used in an appropriate stoichiometric calculation to derive the mass of the relevant element.
mol C = 0.00394 g CO 2 × 44.01 g/mol 1 mol CO^2 × (^) 1 mol CO1 mol C 2
= 8.95 × 10−5^ mol C
mol H = 0.00161 g H 2 O ×
1 mol H 2 O 18.02 g/mol ×^
2 mol H 1 mol H 2 O = 1.79 × 10
−4 (^) mol H
The empirical formula for the compound is then derived by identifying the smallest whole-number multiples for these molar amounts. The H-to-C molar ratio is
mol H mol C =
1.79 × 10−4^ mol H 8.95 × 10−5^ mol C
= 2 mol H1 mol C
and the empirical formula for polyethylene is CH 2.
Check Your Learning
A 0.00215-g sample of polystyrene, a polymer composed of carbon and hydrogen, produced 0.00726 g of CO 2 and 0.00148 g of H 2 O in a combustion analysis. What is the empirical formula for polystyrene?
Answer: CH
4.5 Quantitative Chemical Analysis
78. What volume of 0.0105-M HBr solution is be required to titrate 125 mL of a 0.0100- M Ca(OH) 2 solution? Ca(OH) 2 ( aq ) + 2HBr( aq ) ⟶ CaBr 2 ( aq ) + 2H 2 O( l ) 79. Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidity of the rain is due to the presence of sulfuric acid, what was the concentration of sulfuric acid in this sample of rain? 80. What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO 3 requires 20.22 mL of the AgNO 3 solution to reach the end point? AgNO 3 ( aq ) + NaCl( aq ) ⟶ AgCl( s ) + NaNO 3 ( aq ) 81. In a common medical laboratory determination of the concentration of free chloride ion in blood serum, a serum sample is titrated with a Hg(NO 3 ) 2 solution. 2Cl−( aq ) + Hg(NO 3 ) 2 ( aq ) ⟶ 2NO 3 −( aq ) + HgCl 2 ( s )
What is the Cl−^ concentration in a 0.25-mL sample of normal serum that requires 1.46 mL of 5.25 × 10 −4^ M Hg(NO 3 ) 2 ( aq ) to reach the end point?
82. Potatoes can be peeled commercially by soaking them in a 3-M to 6-M solution of sodium hydroxide, then removing the loosened skins by spraying them with water. Does a sodium hydroxide solution have a suitable concentration if titration of 12.00 mL of the solution requires 30.6 mL of 1.65 M HCI to reach the end point? 83. A sample of gallium bromide, GaBr 2 , weighing 0.165 g was dissolved in water and treated with silver nitrate,
AgNO 3 , resulting in the precipitation of 0.299 g AgBr. Use these data to compute the %Ga (by mass) GaBr 2.
84. The principal component of mothballs is naphthalene, a compound with a molecular mass of about 130 amu, containing only carbon and hydrogen. A 3.000-mg sample of naphthalene burns to give 10.3 mg of CO 2. Determine
its empirical and molecular formulas.
85. A 0.025-g sample of a compound composed of boron and hydrogen, with a molecular mass of ~28 amu, burns spontaneously when exposed to air, producing 0.063 g of B 2 O 3. What are the empirical and molecular formulas of
the compound.
86. Sodium bicarbonate (baking soda), NaHCO 3 , can be purified by dissolving it in hot water (60 °C), filtering to remove insoluble impurities, cooling to 0 °C to precipitate solid NaHCO 3 , and then filtering to remove the solid,
leaving soluble impurities in solution. Any NaHCO 3 that remains in solution is not recovered. The solubility of
NaHCO 3 in hot water of 60 °C is 164 g L. Its solubility in cold water of 0 °C is 69 g/L. What is the percent yield of NaHCO 3 when it is purified by this method?
87. What volume of 0.600 M HCl is required to react completely with 2.50 g of sodium hydrogen carbonate? NaHCO 3 ( aq ) + HCl( aq ) ⟶ NaCl( aq ) + CO 2 ( g ) + H 2 O( l ) 88. What volume of 0.08892 M HNO 3 is required to react completely with 0.2352 g of potassium hydrogen
phosphate? 2HNO 3 ( aq ) + K 2 HPO 4 ( aq ) ⟶ H 2 PO 4 ( aq ) + 2KNO 3 ( aq )
89. What volume of a 0.3300- M solution of sodium hydroxide would be required to titrate 15.00 mL of 0.1500 M oxalic acid? C 2 O 4 H 2 ( aq ) + 2NaOH( aq ) ⟶ Na 2 C 2 O 4 ( aq ) + 2H 2 O( l )
