Question

Consider the titration of \(100.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\) \(\left(K_{\mathrm{b}}=3.0 \times 10^{-6}\right)\) 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. \(0.0 \mathrm{~mL}\) b. \(20.0 \mathrm{~mL}\) c. \(25.0 \mathrm{~mL}\) d. \(40.0 \mathrm{~mL}\) e. \(50.0 \mathrm{~mL}\) f. \(100.0 \mathrm{~mL}\)

Short answer

The short answer for each volume of HNO3 is as follows: a. pH = 11.19 b. pH = 10.63 c. pH = 10.44 d. pH = 10.09 e. pH = 1.56 f. pH = 1.30

Step-by-step solution

List the given quantities and constants

- Volumes: \(100.0 \ \text{mL} \text{ H}_{2}\text{NNH}_{2},\) volumes of \(\text{HNO}_{3}\) are given in the problem. - Concentrations: \(0.100 \ \text{M H}_{2}\text{NNH}_{2}, 0.200 \ \text{M HNO}_{3}\) - Equilibrium constant: \(K_\text{b}=3.0 \times 10^{-6}\)

Set up a general formula for the amount of base and acid at each point

- Moles of base, \(\text{H}_{2}\text{NNH}_{2}: moles_{base} = initial\ base\ concentration \times initial\ base\ volume - \frac{acid\ volume \times acid\ concentration}{2}\) - Moles of acid, \(\text{HNO}_3: moles_{acid} = acid\ volume \times acid\ concentration\)

Calculate moles of the species for given volumes of HNO3 and then the resulting concentrations

For each volume, calculate the moles of species and then the resulting concentrations. The moles and concentrations of the species will be different at each volume. a. \(0.0\ \text{mL HNO}_3\) b. \(20.0\ \text{mL HNO}_3\) c. \(25.0\ \text{mL HNO}_3\) d. \(40.0\ \text{mL HNO}_3\) e. \(50.0\ \text{mL HNO}_3\) f. \(100.0\ \text{mL HNO}_3\)

Calculate the pH of the resulting solutions

For each case, calculate the pH using the appropriate formulas and relations between the equilibrium constant and the concentrations of the species present. a. The added acid volume is zero, so the solution is just the base with no acid. Calculate the initial pOH using the Kb value and then find the pH by subtracting it from 14. b. and c. This is the buffer region, use the Henderson-Hasselbalch equation to calculate pH: \[pH = pK_a + \log{\frac{[\text{A}^-]}{[\text{HA}]}}\] d. This is the equivalence point, the base and acid have reacted completely. Calculate the pH using only the conjugate species that are produced in the reaction. e. and f. This is the excess acid region, calculate pH directly by taking the negative logarithm of the H+/H3O+ ion concentration in the solution.

Key concepts

Acid-Base Reaction

An acid-base reaction occurs when an acid reacts with a base to produce water and a salt. This process is a fundamental concept in chemistry known as neutralization. In the given problem, the base H\(_2\)NNH\(_2\) is titrated with the acid HNO\(_3\). When these two react, they neutralize each other forming water and a salt — in this case, the salt is a nitrate of H\(_2\)NNH\(_2\). Understanding the initial concentration and volume of both the acid and the base is critical because it determines how much product is formed and whether there will be any excess acid or base after the reaction completes.
This reaction involves proton transfer, where the base H\(_2\)NNH\(_2\) accepts protons from the acid HNO\(_3\). This exchange of protons is what alters the pH of the solution at different stages of the titration process.

pH Calculation

Calculating the pH of a solution involves determining the hydrogen ion (H\(^+\)) concentration. For bases, we often find the pOH first and then convert it to pH because bases deal with hydroxide ions (OH\(^-\)). The relationship between pH and pOH is expressed as:

• pH + pOH = 14

In the starting solution with no acid added, we use the given K\(_b\) constant to find the [OH\(^-\)] concentration. From there, we calculate the pOH and finally obtain the pH.
During titration, with small additions of acid, you may need to apply the Henderson-Hasselbalch equation because it simplifies pH calculations for buffering solutions:

• pH = pK\(_a\) + \(\log{\frac{[ ext{A}^-]}{[ ext{HA}]}}\)

It is crucial to realize how pH changes as more acid is added: moving from a basic (high pH ) solution towards acidic (low pH ). Each point in the titration curve presents a different state of the reaction, requiring specific pH calculation methods.

Equivalence Point

The equivalence point in a titration is the stage where the amount of acid equals the amount of base, resulting in complete neutralization. At this point, the number of moles of the acid added equals the number of moles of the base present initially. Thus, the resulting solution’s pH depends solely on the species produced from the neutralization.
In the problem at hand, reaching the equivalence point means the entire initial moles of H\(_2\)NNH\(_2\) have reacted with the equivalent moles of HNO\(_3\) to form its conjugate acid, which impacts pH. The resulting solution’s pH might not be neutral (pH = 7) due to the conjugate acid’s and base’s respective strengths, a concept important when dealing with weak acids or weak bases. Calculating the pH at the equivalence point requires considering this newly formed species and its dissociation in water.

Buffer Solution

A buffer solution resists drastic changes in pH upon the addition of small amounts of acid or base. It is typically composed of a weak acid and its conjugate base or a weak base and its conjugate acid. Within the provided titration example, the pH regions during the buffer zones are calculated using the Henderson-Hasselbalch equation.
This buffer effect is particularly important at the beginning and intermediate stages of the titration where the solution still contains significant amounts of both reactants and products (unreacted base and conjugate acid).

• pH = pK\(_a\) + \(\log{\frac{[ ext{A}^-]}{[ ext{HA}]}}\)

Applying this formula helps in determining the pH of the system accurately, showcasing the buffer’s capacity to stabilize the solution's pH against minor changes. Understanding buffer solutions is essential because they play a critical role in many biological and chemical systems where maintaining a stable pH is crucial.