Question

How many milliliters of \(0.0850 \mathrm{M} \mathrm{NaOH}\) are required to titrate each of the following solutions to the equivalence point: (a) \(40.0 \mathrm{~mL}\) of \(0.0900 \mathrm{M} \mathrm{HNO}_{3}\), (b) \(35.0 \mathrm{~mL}\) of \(0.0850 \mathrm{MCH}_{3} \mathrm{COOH}\), (c) \(50.0 \mathrm{~mL}\) of a solution that contains \(1.85 \mathrm{~g}\) of \(\mathrm{HCl}\) per liter?

Short answer

The required volume of \(0.0850 \mathrm{M} \mathrm{NaOH}\) to titrate each solution at their equivalence point is: (a) \(42.4 \mathrm{~mL}\) (b) \(35.1 \mathrm{~mL}\) (c) \(29.9 \mathrm{~mL}\)

Step-by-step solution

Determine the moles of the given solutions.

To calculate the number of moles of each solution, we can use the formula: Moles = molarity × volume Here, we must convert the given volume of each solution into liters before multiplying it by its molarity.

Calculate the required moles of NaOH for each reaction.

For each given solution, we must find the amount of NaOH needed to reach the equivalence point. The basis behind this calculation is the stoichiometry, assuming a 1:1 ratio (monoprotic species). We can use the determined moles of the given solution from step 1 as the required moles of NaOH.

Determine the volume of NaOH needed for each reaction.

Now, we have the required moles of NaOH for each solution. To find the volume, we can use the given molarity of NaOH, \(0.0850 \mathrm{M}\), and the formula: Volume = moles / molarity (a) \(40.0 \mathrm{~mL}\) of \(0.0900 \mathrm{M} \mathrm{HNO}_{3}\):

(a) Step 1: Moles of HNO3

Moles of HNO3 = \((40.0 \mathrm{~mL} \times 0.001 \frac{\mathrm{L}}{\mathrm{mL}}) \times 0.0900 \mathrm{M} = 0.00360 \mathrm{~mol}\)

(a) Step 2: Moles of NaOH

Moles of NaOH = \(0.00360 \mathrm{~mol}\) (1:1 stoichiometry)

(a) Step 3: Volume of NaOH

Volume of NaOH = \(\frac{0.00360 \mathrm{~mol}}{0.0850 \mathrm{M}} = 0.0424 \mathrm{~L} = 42.4 \mathrm{~mL}\) (b) \(35.0 \mathrm{~mL}\) of \(0.0850 \mathrm{MCH}_{3} \mathrm{COOH}\):

(b) Step 1: Moles of CH3COOH

Moles of CH3COOH = \((35.0 \mathrm{~mL} \times 0.001 \frac{\mathrm{L}}{\mathrm{mL}}) \times 0.0850 \mathrm{M} = 0.00298 \mathrm{~mol}\)

(b) Step 2: Moles of NaOH

Moles of NaOH = \(0.00298 \mathrm{~mol}\) (1:1 stoichiometry)

(b) Step 3: Volume of NaOH

Volume of NaOH = \(\frac{0.00298 \mathrm{~mol}}{0.0850 \mathrm{M}} = 0.0351 \mathrm{~L} = 35.1 \mathrm{~mL}\) (c) \(50.0 \mathrm{~mL}\) of a solution that contains \(1.85 \mathrm{~g}\) of \(\mathrm{HCl}\) per liter:

(c) Step 1: Calculate molarity of HCl

Molar mass of HCl = \(1.008 \mathrm{~g/mol} + 35.45 \mathrm{~g/mol} = 36.458 \mathrm{~g/mol}\) Molarity of HCl = \(\frac{1.85 \mathrm{~g/L}}{36.458 \mathrm{~g/mol}} = 0.0507 \mathrm{M}\)

(c) Step 2: Moles of HCl

Moles of HCl = \((50.0 \mathrm{~mL} \times 0.001 \frac{\mathrm{L}}{\mathrm{mL}}) \times 0.0507 \mathrm{M} = 0.00254 \mathrm{~mol}\)

(c) Step 3: Moles of NaOH

Moles of NaOH = \(0.00254 \mathrm{~mol}\) (1:1 stoichiometry)

(c) Step 4: Volume of NaOH

Volume of NaOH = \(\frac{0.00254 \mathrm{~mol}}{0.0850 \mathrm{M}} = 0.0299 \mathrm{~L} = 29.9 \mathrm{~mL}\) In conclusion, the required volume of \(0.0850 \mathrm{M} \mathrm{NaOH}\) to titrate each solution at their equivalence point is as follows: (a) \(42.4 \mathrm{~mL}\) (b) \(35.1 \mathrm{~mL}\) (c) \(29.9 \mathrm{~mL}\)

Key concepts

Stoichiometry

Stoichiometry is akin to a recipe for chemistry. It involves the calculation of the quantities of reactants and products involved in a chemical reaction. It's based on the conservation of mass and the concept of moles, which allows chemists to describe amounts of substances. A balanced chemical equation is the starting point for all stoichiometry calculations because it reflects the necessary proportions of all reactants and products.

In the case of acid-base titrations, stoichiometry helps to determine the exact amount of titrant (in this scenario, sodium hydroxide, NaOH) needed to completely neutralize the analyte (the acid in the solution). Typically, acid-base reactions have straightforward stoichiometric ratios. For monoprotic acids (like HNO3 and CH3COOH in our examples), the ratio is often 1:1 with the base (NaOH). By understanding and applying stoichiometry, students not only perform calculations more effectively but also gain a deeper understanding of the underlying chemical processes.

Molarity Calculation

Molarity, symbolized as M, is one of the most commonly used units of concentration in chemistry. It quantifies the number of moles of a solute per liter of solution. The formula for molarity is:
\[ Molarity = \frac{moles \text{ of solute}}{liters \text{ of solution}} \]

When calculating molarity, it's essential to ensure that volumes are converted to liters since molarity is defined per liter. In titration problems, molarity plays a critical role because it allows us to calculate the number of moles of the titrant that will be needed to reach the equivalence point. Once we know the number of moles and the molarity of the titrant, it becomes a straightforward step to find out the volume required to neutralize the given solution by rearranging the molarity formula.

Equivalence Point

The equivalence point in a titration is a crucial concept and marks the moment when the amount of titrant added is stoichiometrically equivalent to the amount of substance present in the analyte. At this point, the moles of acid and base in solution are equal, and the acid has been completely neutralized by the base, or vice versa, depending on the direction of titration. In our examples, the equivalence point is achieved when the moles of NaOH added match the moles of acid present in the sample.

The equivalence point is often signaled by a sudden change in some property of the solution, such as pH, and is frequently indicated by a color change if an appropriate indicator has been added. In educational settings, understanding the concept of the equivalence point helps students to grasp the end goal of a titration and, through suitable calculations, allows for accurate determination of unknown concentrations in solution.