Question

Sodium hydroxide is used extensively in acid-base titrations because it is a strong, inexpensive base. A sodium hydroxide solution was standardized by titrating \(25.00 \mathrm{~mL}\) of \(0.1528 \mathrm{M}\) standard hydrochloric acid. The initial buret reading of the sodium hydroxide was \(2.24 \mathrm{~mL},\) and the final reading was \(39.21 \mathrm{~mL}\). What was the molarity of the base solution?

Short answer

The molarity of the NaOH solution is 0.1034 M.

Step-by-step solution

- Determine the volume of NaOH used

To find the volume of NaOH solution used in the titration, subtract the initial buret reading from the final buret reading: \[\text{{Volume of NaOH used}} = \text{{Final buret reading}} - \text{{Initial buret reading}} \ = 39.21 \text{ mL} - 2.24 \text{ mL} = 36.97 \text{ mL}\]

- Convert the volume of NaOH to liters

Since molarity is expressed in mol/L, convert the volume from milliliters to liters: \[\text{{Volume of NaOH in liters}} = \frac{36.97 \text{ mL}}{1000} = 0.03697 \text{ L}\]

- Find the moles of HCl used

Use the molarity and volume of the HCl solution to calculate the moles of HCl that reacted: \[\text{{Moles of HCl}} = \text{{Molarity of HCl}} \times \text{{Volume of HCl in L}} = 0.1528 \text{ M} \times 0.02500 \text{ L} = 0.00382 \text{ moles} \]

- Determine the moles of NaOH required to neutralize HCl

In this reaction, sodium hydroxide (NaOH) reacts with hydrochloric acid (HCl) in a 1:1 molar ratio: \[\text{{Moles of NaOH}} = \text{{Moles of HCl}} = 0.00382 \]

- Calculate the molarity of NaOH solution

Molarity is defined as moles of solute per liter of solution. Thus, we find the molarity of the NaOH solution: \[\text{{Molarity of NaOH}} = \frac{{\text{{Moles of NaOH}}}}{\text{{Volume of NaOH in liters}}} = \frac{0.00382 \text{ moles}}{0.03697 \text{ L}} = 0.1034 \text{ M} \]

Key concepts

Molarity Calculation

Molarity is a measure of concentration in terms of moles of solute per liter of solution. In titrations, solutions' molarity helps determine the amount of reactants. To calculate molarity, remember this key formula: \( \text{Molarity} (M) = \frac{\text{Moles of Solute}}{\text{Volume of Solution in Liters}} \). Here, you need to know the moles of solute, which are often given or found using the molar mass and volume given. Also, don’t forget to convert volumes to liters if they are given in milliliters. In our problem, \( 0.00382 \text{ moles HCl} \) reacted with an unknown NaOH volume, which helped determine the NaOH’s molarity.

Neutralization Reaction

Neutralization involves an acid and base reacting to form water and a salt. This reaction always follows the basic formula: \( \text{Acid + Base} \rightarrow \text{Water + Salt} \). If the concentrations (molarity) and volumes of reactants are known, we can use stoichiometry to find unknown concentrations. In titration, since we use a strong acid and strong base (HCl and NaOH), they react in a 1:1 molar ratio. This ratio makes it straightforward to calculate one reactant's required moles given the moles of another. In our example, each mole of HCl neutralizes one mole of NaOH.

Volumetric Analysis

Volumetric analysis involves measuring the volume of a solution to determine its concentration and quantity of solute. In titrations, a solution of known concentration (titrant) is added to a solution of unknown concentration until the reaction is complete. The point of completion is usually indicated by a color change (from an indicator) or an electrical measurement. Here, we measured the initial and final buret readings to determine the volume of NaOH used. For HCl of known molarity, the exact volume reacted helped in calculating the NaOH's concentration.

Stoichiometry

Stoichiometry is the study of mass and molar relationships between reactants and products in a chemical reaction. To solve stoichiometric problems, we often use balanced chemical equations to convert between moles of different substances. For example, if we know the moles of one reactant, we can calculate the moles of another reactant or product using the coefficients from the balanced equation. In our titration, because we had a 1:1 molar ratio between HCl and NaOH, the moles of HCl gave a direct measure of the moles of NaOH needed for complete neutralization. From there, we calculated the NaOH molarity accurately.