Question

Consider the titration of \(60.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{HNO}_{3}\) with \(0.450 \mathrm{M}\) \(\mathrm{NaOH}\). (a) How many millimoles of \(\mathrm{HNO}_{3}\) are present at the start of the titration? (b) How many milliliters of \(\mathrm{NaOH}\) are required to reach the equivalence point? (c) What is the \(\mathrm{pH}\) at the equivalence point? (d) Sketch the general shape of the pH titration curve.

Short answer

a) 9 millimoles HNO₃. b) 20 mL NaOH. c) pH = 7.

Step-by-step solution

Calculate Initial Millimoles of HNO₃

To find the initial millimoles of HNO₃, use the formula: \( ext{millimoles} = Molarity \times Volume \). The volume should be in liters. Here, \( M = 0.150 \) M and \( V = 60.0 \) mL = \( 0.060 \) L. So, the millimoles of HNO₃ is: \( 0.150 \times 0.060 = 0.009 \, \text{mol} \) or 9 millimoles.

Determine Volume of NaOH for Equivalence Point

At the equivalence point, the moles of \(\mathrm{NaOH}\) added will equal the initial moles of \(\mathrm{HNO}_{3}\). Since we have 9 millimoles of \(\mathrm{HNO}_{3}\), we need 9 millimoles of \(\mathrm{NaOH}\). The molarity of \(\mathrm{NaOH}\) is 0.450 M, so the volume needed is: \( V = \frac{9 \text{ mmol}}{0.450 \text{ M}} = 20 \text{ mL} \).

Calculate pH at the Equivalence Point

At the equivalence point of a strong acid-strong base titration, the solution consists only of the salt and water. For \(\mathrm{HNO}_{3} + \mathrm{NaOH} = \mathrm{NaNO}_{3} + \mathrm{H}_{2}\mathrm{O}\), the pH is neutral (pH = 7) because this is a strong acid/strong base titration.

Sketch the Titration Curve

The titration curve begins at a low pH, as the initial solution is an acidic solution of \(\mathrm{HNO}_{3}\). As \(\mathrm{NaOH}\) is added, the pH gradually rises. At the equivalence point, the pH suddenly jumps to 7 due to complete neutralization. Beyond the equivalence point, the pH increases sharply as extra \(\mathrm{NaOH}\) makes the solution basic.

Key concepts

Equivalence Point

The equivalence point in a titration is a critical stage where the amount of titrant added is just enough to completely neutralize the analyte solution. In our example with - \(\mathrm{HNO}_{3} \) and \(\mathrm{NaOH}\), the equivalence point is achieved when the moles of \(\mathrm{NaOH}\) equal the moles of \(\mathrm{HNO}_{3}\).- It indicates that the original substance (\(\mathrm{HNO}_{3}\)) in the solution has been fully consumed. Reaching this point implies complete neutralization for a strong acid-strong base titration like ours. It is the moment where the chemical reaction is perfectly balanced, leaving only the resulting salt and water in the solution. Identifying the equivalence point is crucial since it tells us that the titration is accurately complete. This stage is often characterized by a noticeable change in some property of the solution, such as its pH level.

Strong Acid-Strong Base Titration

Titrations involving strong acids and strong bases, such as \(\mathrm{HNO}_{3}\) and \(\mathrm{NaOH}\), follow a straightforward neutralization reaction because both species fully dissociate in water.- The reaction between \(\mathrm{HNO}_{3}\), a strong acid, and \(\mathrm{NaOH}\), a strong base, goes to completion.- This reaction can be represented as follows: \(\mathrm{HNO}_{3} + \mathrm{NaOH} \rightarrow \mathrm{NaNO}_{3} + \mathrm{H}_{2}\mathrm{O}\).
Strong acid-strong base titrations are typically characterized by sharp changes in pH near the equivalence point. In practice, these titrations are quite useful in determining the concentration of an unknown solution since the complete dissociation ensures a precise stoichiometric relationship. The final product, a salt and water, typically doesn't affect the pH significantly at the equivalence point, which is why it's neutral in these conditions.

pH Calculation

Calculating the pH at various points during a titration is instrumental in understanding the titration curve, especially its steep rise and sudden jumps.- Initially, the solution's pH is determined by the concentration of the strong acid (\(\mathrm{HNO}_{3}\)), which is low as its dissociation in water produces a high concentration of \(\mathrm{H}^{+}\) ions. - As \(\mathrm{NaOH}\) is added, it neutralizes the \(\mathrm{H}^{+}\) ions, causing the pH to rise gradually.
At the equivalence point in a strong acid-strong base titration, the pH is neutral, usually around 7.- Past the equivalence point, the pH is dictated by the excess strong base in the solution, causing a sharp increase in pH. Accurately calculating pH values at these points helps sketch the titration curve and understand the acid-base behavior in the solution.

Titration Curve

A titration curve graphically represents the change in pH during a titration process, providing a visual tool for identifying key stages like the equivalence point. - The curve for a strong acid-strong base titration starts at a low pH due to the initial acidic nature of the solution.- As the titrant (\(\mathrm{NaOH}\)) is slowly added, the pH increases steadily until reaching a significant upward spike.
- This steep increase is indicative of reaching the equivalence point, where the solution has been completely neutralized, typically reflected by a vertical curve section around pH 7. - Beyond this point, as more \(\mathrm{NaOH}\) is added, the solution becomes basic, and the curve continues to rise at a moderate slope. Titration curves are essential for precisely determining when neutralization occurs and examining the characteristics and behavior of the substances involved.

Millimoles Calculation

Calculating millimoles (mmol) is a preliminary and crucial step in titration problems to ensure correct proportions and interactions. - Millimoles are calculated using the formula: \(\text{millimoles} = \text{Molarity} \times \text{Volume}\). In this case, ensure the volume is converted to liters.- For instance, with our example, we calculated that the \(\mathrm{HNO}_{3}\) solution has 9 millimoles initially (\(0.150\,\text{M} \times 0.060\,\text{L} = 0.009\,\text{mol}\) or 9 mmol). - Understanding the initial quantity helps us calculate how much of the titrant (\(\mathrm{NaOH}\)) is required to reach the equivalence point.Together, these calculations form the backbone of understanding titration processes, allowing one to analyze and predict the behavior of chemical reactions quantitatively.