Question

How many milliliters of \(0.105 \mathrm{M} \mathrm{HCl}\) are needed to titrate each of the following solutions to the equivalence point: (a) \(45.0 \mathrm{~mL}\) of \(0.0950 \mathrm{M} \mathrm{NaOH}\), (b) \(22.5 \mathrm{~mL}\) of \(0.118 \mathrm{M} \mathrm{NH}_{3}\), (c) \(125.0 \mathrm{~mL}\) of a solution that contains \(1.35 \mathrm{~g}\) of \(\mathrm{NaOH}\) per liter?

Short answer

The volumes of \(0.105\mathrm{~M}\mathrm{HCl}\) needed to titrate the given solutions to the equivalence point are: (a) 40.5 mL for \(45.0\mathrm{~mL}\) of \(0.0950\mathrm{M}\mathrm{NaOH}\), (b) 25.5 mL for \(22.5\mathrm{~mL}\) of \(0.118\mathrm{~M}\mathrm{NH}_3\), and (c) 40.0 mL for \(125.0\mathrm{~mL}\) of a solution containing \(1.35\mathrm{~g}/\mathrm{L}\) \(\mathrm{NaOH}\).

Step-by-step solution

(a) Titrate \(45.0\mathrm{~mL}\) of \(0.0950\mathrm{M} \mathrm{NaOH}\)

Given: \(M_{\mathrm{HCl}} = 0.105\mathrm{~M}\) \(M_{\mathrm{NaOH}} = 0.0950\mathrm{~M}\) \(V_{\mathrm{NaOH}} = 45.0\mathrm{~mL}\) Find \(V_{\mathrm{HCl}}\) \(V_{\mathrm{HCl}} = \frac{M_{\mathrm{NaOH}}×V_{\mathrm{NaOH}}}{M_{\mathrm{HCl}}}\) \(V_{\mathrm{HCl}} = \frac{0.0950\mathrm{~M} × 45.0\mathrm{~mL}}{0.105\mathrm{~M}}\) \(V_{\mathrm{HCl}} = 40.5\mathrm{~mL}\) Therefore, 40.5 mL of \(0.105\mathrm{~M}\mathrm{HCl}\) are needed to titrate the \(45.0\mathrm{~mL}\) of \(0.0950\mathrm{M}\mathrm{NaOH}\).

(b) Titrate \(22.5\mathrm{~mL}\) of \(0.118\mathrm{M}\mathrm{NH}_3\)

Given: \(M_{\mathrm{NH}_3} = 0.118\mathrm{~M}\) \(V_{\mathrm{NH}_3} = 22.5\mathrm{~mL}\) Find \(V_{\mathrm{HCl}}\) \(V_{\mathrm{HCl}} = \frac{M_{\mathrm{NH}_3}×V_{\mathrm{NH}_3}}{M_{\mathrm{HCl}}}\) \(V_{\mathrm{HCl}} = \frac{0.118\mathrm{~M} × 22.5\mathrm{~mL}}{0.105\mathrm{~M}}\) \(V_{\mathrm{HCl}} = 25.5\mathrm{~mL}\) Therefore, 25.5 mL of \(0.105\mathrm{~M}\mathrm{HCl}\) are needed to titrate the \(22.5\mathrm{~mL}\) of \(0.118\mathrm{~M}\mathrm{NH}_3\).

(c) Titrate \(125.0\mathrm{~mL}\) of solution containing \(1.35\mathrm{~g}/\mathrm{L}\) \(\mathrm{NaOH}\)

First, we need to find the molarity of the \(\mathrm{NaOH}\) solution: \(1.35 \frac{\mathrm{g}}{\mathrm{L}} \times \frac{1 \mathrm{L}}{1000 \mathrm{~mL}} \times \frac{1 \mathrm{mol}}{40.0 \mathrm{~g}} = 0.03375 \mathrm{~M}\) Given: \(M_{\mathrm{NaOH}} = 0.03375\mathrm{~M}\) \(V_{\mathrm{NaOH}} = 125.0\mathrm{~mL}\) Find \(V_{\mathrm{HCl}}\) \(V_{\mathrm{HCl}} = \frac{M_{\mathrm{NaOH}}×V_{\mathrm{NaOH}}}{M_{\mathrm{HCl}}}\) \(V_{\mathrm{HCl}} = \frac{0.03375\mathrm{ M} × 125.0\mathrm{~mL}}{0.105\mathrm{~M}}\) \(V_{\mathrm{HCl}} = 40.0\mathrm{~mL}\) Therefore, 40.0 mL of \(0.105\mathrm{ M}\mathrm{HCl}\) are needed to titrate the \(125.0\mathrm{~mL}\) of solution containing \(1.35\mathrm{~g}/\mathrm{L}\) \(\mathrm{NaOH}\).

Key concepts

Understanding Molarity

Molarity, represented as 'M', is a measure of concentration that tells us how many moles of a given substance are present in one liter of solution. It's a fundamental concept in chemistry used in various calculations, including those concerning solutions and reactions. To calculate the molarity of a solution, one must know the amount of solute (in moles) and the volume of the solution (in liters). The formula for molarity is defined as:
\[ Molarity (M) = \frac{moles \, of \, solute}{liters \, of \, solution} \]
For example, if we dissolve 2 moles of sodium chloride (NaCl) in enough water to make a 1-liter solution, the molarity of that solution is 2 M. When solving titration problems like the one in the exercise, molarity helps us determine the volume of titrant needed to reach the equivalence point, where the number of moles of acid equates to the number of moles of base, achieving neutralization.

The Role of Stoichiometry in Titration

Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in a chemical reaction. It's based on the conservation of mass and the law of definite proportions. Stoichiometry allows chemists to predict the outcomes of reactions, determine the proportions of reactants needed, and calculate product yields.
When performing a titration, stoichiometry is used to relate the volume and molarity of one solution to the volume and molarity of the other through the balanced chemical equation. For neutralization reactions, the stoichiometry usually involves a 1:1 molar ratio between acid and base when they are strong, but can vary for weak acids or bases. For the textbook problem, we employ the concepts of stoichiometry to calculate the volume of the hydrochloric acid (HCl) solution needed to neutralize the given amount of sodium hydroxide (NaOH) or ammonia (NH3). This calculation hinges on the moles of HCl being equal to the moles of the other substance at the equivalence point, as per the reaction's stoichiometry.

Neutralization Reaction Explained

A neutralization reaction is a chemical reaction in which an acid and a base react quantitatively to form water and a salt. This reaction is central to acid-base titration exercises like the one provided. The general formula for a neutralization reaction is:
\[ Acid + Base \rightarrow Water + Salt \]
These reactions proceed until the number of moles of acid equals the number of moles of the base, known as the equivalence point. To reach this point in a titration, a known concentration of one reactant (titrant) is gradually added to a known volume of another reactant until neutralization is achieved. In the textbook problem, we use the molarity of both the acid (HCl) and the base (NaOH or NH3) to calculate the volume of acid needed to reach the equivalence point. By understanding the concept of neutralization, we can interpret titration data, such as pH changes, and perform accurate calculations in analytical chemistry.