Question

Three students titrate different samples of the same solution of \(\mathrm{HCl}\) to obtain its molarity. Below are their data. Student \(\mathrm{A}: \quad 20.00 \mathrm{~mL} \mathrm{HCl}+20.00 \mathrm{~mL} \mathrm{H}_{2} \mathrm{O}\) \(0.100 \mathrm{M} \mathrm{NaOH}\) used to titrate to the equivalence point Student B: \(\quad 20.00 \mathrm{~mL} \mathrm{HCl}+40.00 \mathrm{~mL} \mathrm{H}_{2} \mathrm{O}\) \(0.100 \mathrm{M} \mathrm{NaOH}\) used to titrate to the equivalence point Student C: \(\quad 20.00 \mathrm{~mL} \mathrm{HCl}+20.00 \mathrm{~mL} \mathrm{H}_{2} \mathrm{O}\) \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) used to titrate to the equivalence point. All the students calculated the molarities correctly. Which (if any) of the following statements are true? (a) The molarity calculated by \(A\) is half that calculated by \(B\). (b) The molarity calculated by \(\mathrm{A}\) is equal to that calculated by \(\mathrm{C}\). (c) The molarity calculated by B is twice that calculated by C. (d) The molarity calculated by \(\mathrm{A}\) is twice that calculated by \(\mathrm{B}\). (e) The molarity calculated by \(\mathrm{A}\) is equal to that calculated by \(\mathrm{B}\). Challenge Problems

Short answer

Answer: (e) The molarities determined by students A and B using NaOH as a titrant are equal, as they both used the same amount of HCl and NaOH, resulting in the same molarity calculations.

Step-by-step solution

Determine the moles of HCl in each titration

Since all students are using the same HCl solution, we know that the number of moles of HCl will be the same for all three titrations. Therefore, the moles of HCl can be expressed as: moles of HCl = Molarity × Volume

Relate the moles of titrant to the moles of HCl

For students A and B, they used NaOH as the titrant. The reaction between HCl and NaOH is 1:1, so the moles of NaOH are equal to the moles of HCl. For student C, they used Ba(OH)2 as the titrant. The reaction between HCl and Ba(OH)2 is 2:1, so the moles of Ba(OH)2 will be equal to half the moles of HCl.

Find the molarities calculated by A, B, and C

Using the moles of titrant and the volume of HCl, we can calculate the molarities of HCl for each student. For student A: M1 = (Moles of NaOH) / (20.00 mL HCl) M1 = (20.00 × 0.100 mol/L) / 20.00 For student B: M2 = (Moles of NaOH) / (20.00 mL HCl) M2 = (20.00 × 0.100 mol/L) / 20.00 For student C, note that moles of Ba(OH)2 = 0.5 × moles of HCl: M3 = (2 × Moles of Ba(OH)2) / (20.00 mL HCl) M3 = 2 × (20.00 × 0.100 mol/L) / 20.00

Assess the truth of statements (a) to (e)

(a) False, since M1 and M2 have the same molarity. (b) False, since M1 is only half of M3. (c) False, since M2 is only half of M3. (d) False, since M1 and M2 have the same molarity. (e) True, since M1 and M2 have the same molarity. Thus, the only true statement is (e).

Key concepts

Molarity Calculation

Understanding how to calculate molarity is crucial in chemistry, especially when dealing with solutions and reactions. Molarity, denoted by the capital letter M, is a measure of the concentration of a solute in a solution. It's defined as the number of moles of solute divided by the volume of the solution in liters. Mathematically, the formula for molarity (M) is expressed as:
\[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
In the context of an acid-base titration, molarity calculation becomes particularly important for figuring out how concentrated the acid or base is. For instance, if a student titrates a sample of hydrochloric acid (HCl) with a sodium hydroxide (NaOH) solution of known molarity, they can find the molarity of the acid by rearranging the molarity formula:
\[ \text{Molarity of HCl} = \frac{\text{Moles of NaOH}}{\text{Volume of HCl in liters}} \]
When calculating the molarity for the given exercise, you would consider the volume of only the HCl solution — not the total volume after adding water — because water does not change the number of moles of HCl present. It's important to note that dilution with water does not affect the molarity calculation in titrations since the molarity is defined for a specific volume of the solution being titrated, regardless of any additional solvent added.

Stoichiometry

Stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction. It's all about the math behind chemical equations, and knowing stoichiometry allows chemists to predict how much of each substance is needed or produced in a reaction. In an acid-base titration, stoichiometry tells us the ratio in which the acid reacts with the base.
For instance, hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH) in a 1:1 stoichiometric ratio, meaning one mole of HCl reacts with one mole of NaOH. However, when HCl reacts with barium hydroxide (Ba(OH)₂), the stoichiometry changes to 2:1 because each mole of Ba(OH)₂ can neutralize two moles of HCl due to it having two hydroxide ions.
\[ \text{For NaOH (Students A and B):} \quad \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
\[ \text{For Ba(OH)}_2\text{(Student C):} \quad 2\text{HCl} + \text{Ba(OH)}_2 \rightarrow \text{BaCl}_2 + 2\text{H}_2\text{O} \]
Applying stoichiometry is vital when figuring out how much titrant to use to reach the equivalence point. By knowing the stoichiometric relationships, you can correctly interpret the results of titrations, as illustrated in the exercise where the stoichiometric ratios directly influenced the calculated molarities.

Equivalence Point

The equivalence point is a key concept in titrations; it represents the point at which the amount of titrant added reacts exactly with the substance being titrated. In a typical acid-base titration, it's the point where the number of moles of acid equals the number of moles of base.
At the equivalence point, the stoichiometry of the reaction dictates that the reactants have combined in their exact ratios, and no excess of either reactant remains. For example, in our exercise involving the titration of HCl with NaOH, the equivalence point was reached once the moles of HCl equaled the moles of NaOH. For Ba(OH)₂, you reach the equivalence point when the moles of HCl are twice the moles of Ba(OH)₂.
Identifying the equivalence point is typically done with the aid of an indicator that changes color at or near this point or by using a pH meter. Accurately determining the equivalence point is crucial for correct molarity calculations, as it ensures that the correct stoichiometric ratio has been used in the calculation. Therefore, in the given exercise, knowing when the equivalence point was reached allowed students to determine the correct molarity of the HCl solution, by accounting for the stoichiometry of the acid-base reaction.