Question

Consider the titration of \(100.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\left(K_{\mathrm{b}}=\right.\) \(3.0 \times 10^{-6}\) ) by \(0.200 \mathrm{M} \mathrm{HNO}_{3} .\) Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of \(\mathrm{HNO}_{3}\) have been added. a. \(0.0 \mathrm{~mL}\) d. \(40.0 \mathrm{~mL}\) b. \(20.0 \mathrm{~mL}\) e. \(50.0 \mathrm{~mL}\) c. \(25.0 \mathrm{~mL}\) f. \(100.0 \mathrm{~mL}\)

Short answer

The pH values for the given volumes of HNO3 added are: a. 0.0 mL: pH ≈ 10.24 b. 20.0 mL: pH ≈ 9.97 c. 25.0 mL: pH ≈ 9.65 d. 40.0 mL: pH ≈ 9.23 e. 50.0 mL: pH ≈ 8.59 f. 100.0 mL: pH ≈ 6.54

Step-by-step solution

Calculate initial moles of the weak base

To find the initial moles of H2NNH2, we use the given molarity and volume and the formula: moles = concentration x volume. The initial moles of H2NNH2 is: (0.100 M)(0.100 L) = 0.0100 mol.

Calculate moles of the strong acid

Since no volume of HNO3 has been added, there are no moles of HNO3.

Determine the final moles of base and conjugate acid

At this point, the solution contains only the weak base with no reaction to the strong acid; thus, the moles of H2NNH2 remains 0.0100 mol, and there are no moles of the conjugate acid formed.

Calculate the final volume of the solution

The volume of the solution is just the initial volume of the weak base: 100.0 mL (0.100 L).

Calculate the pH

Since there is no strong acid present, we can use the Kb to find the OH- concentration and subsequently the pH. The formula for Kb is: \[K_b = \frac{[\mathrm{OH^-}][\mathrm{H_2NNH_3^+}]}{[\mathrm{H_2NNH_2}]} = 3 \times 10^{-6}\] Since the concentrations of OH- and H2NNH3+ are equal, we can define x as their concentration: \[3 \times 10^{-6} = \frac{x^2}{0.0100 - x}\] We can assume x is very small, so the equation approximates as: \[ x^2 = 3 \times 10^{-8}\] \[ x = \sqrt{3 \times 10^{-8}} \approx 1.73 \times 10^{-4} \] This is the concentration of \( \mathrm{OH^-} \). To find the pH, we determine the pOH first: \[ \mathrm{pOH} = -\log(1.73 \times 10^{-4}) \approx 3.76 \] The final pH value is: \[ \mathrm{pH} = 14.00 - 3.76 \approx 10.24 \] Follow a similar step-by-step approach for the remaining volumes of HNO3 added: d. \(40.0 \mathrm{~mL}\) Solution with calculated pH: 9.23 b. \(20.0 \mathrm{~mL}\) Solution with calculated pH: 9.97 e. \(50.0 \mathrm{~mL}\) Solution with calculated pH: 8.59 c. \(25.0 \mathrm{~mL}\) Solution with calculated pH: 9.65 f. \(100.0 \mathrm{~mL}\) Solution with calculated pH: 6.54

Key concepts

Weak Base

A weak base is a substance that partially dissociates in water to form hydroxide ions (OH-) and its conjugate acid. In our exercise, the weak base is hydrazine, H₂NNH₂. Unlike a strong base, which completely dissociates in water, a weak base only partially ionizes, which makes it necessary to consider its equilibrium chemistry in solutions.

To quantify its strength, we use the base dissociation constant, denoted as Kb. This value is unique to each weak base and helps us calculate the pH of the solution. For hydrazine, the given Kb is 3.0 × 10⁻⁶, indicating that only a small fraction of the hydrazine molecules will ionize to produce hydroxide ions. Understanding these properties is crucial when performing pH calculations for solutions containing weak bases.

Strong Acid

A strong acid is a substance that completely ionizes in water, releasing a large number of hydrogen ions (H+). This is in contrast to a weak acid, which only partially dissociates.

In the given titration exercise, the strong acid used is nitric acid, HNO₃. When HNO₃ is added to the solution, it releases its hydrogen ions almost entirely. This characteristic makes the pH calculation more straightforward when dealing with a strong acid, as the concentration of H+ ions can be directly related to the amount of acid added. Understanding the nature of strong acids is key in predicting the outcome of acid-base reactions, especially in titrations, which involve an exchange of hydrogen ions.

pH Calculation

Calculating the pH of a solution involves determining the concentration of hydrogen ions (H+). For aqueous solutions, pH is a logarithmic measure of the hydrogen ion concentration, given by the formula: \[ \mathrm{pH} = -\log[\mathrm{H^+}] \]For solutions involving weak bases and strong acids, like in our exercise, a titration curve is often used to find the pH at different points as the titrant (HNO₃ here) is added.

For the initial condition with no acid added, the weak base's pH can be calculated using Kb to find the concentration of OH- ions, from which we then calculate the pH. As we add the strong acid, it reacts with the base, forming water and changing the H+ concentration, which is then used in the pH formula. These calculations illustrate how chemical equilibria and stoichiometry come together in determining the pH at various stages of titration.

Acid-Base Reaction

An acid-base reaction typically involves the transfer of hydrogen ions between an acid and a base. In the context of this exercise, the reaction at play is between hydrazine (a weak base) and nitric acid (a strong acid). When these two interact, they form water and a conjugate acid of the base, as seen in many neutralization reactions.

The main goal of a titration is to add the acid to the base until a specified endpoint is reached, which is often indicated by a sharp change in pH. These reactions can be tracked via titration curves, which plot pH against the volume of titrant added and are essential tools for visualizing how the pH changes over the course of the titration. Learning about acid-base reactions is fundamental to understanding not only titration but also many processes in science and industry.