Question

An aqueous solution contains dissolved \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\) and \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} .\) The concentration of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) is \(0.50 M\) and \(\mathrm{pH}\) is 4.20 a. Calculate the concentration of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) in this buffer solution. b. Calculate the \(\mathrm{pH}\) after \(4.0 \mathrm{g} \mathrm{NaOH}(s)\) is added to \(1.0 \mathrm{L}\) of this solution. (Neglect any volume change.)

Short answer

The concentration of \(\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{NH}_{3}^{+}\) in the buffer solution is approximately \(0.158 M\). After adding 4.0 g of \(\mathrm{NaOH}\) to the solution, the new pH is approximately 5.78.

Step-by-step solution

Identify the relationship between pH and pKa and the concentrations

We can find the concentration of \(\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{NH}_{3}^{+}\) using the Henderson-Hasselbalch equation, which is given by: \[pH = pKa + \log{\frac{[\mathrm{A^{-}}]}{[\mathrm{HA}]}}\] In our case, the weak base is \(\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{NH}_{2}\) (abbreviated as \(\mathrm{B}\)), and its conjugate acid is \(\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{NH}_{3}^{+}\) (abbreviated as \(\mathrm{BH^{+}}\)). Therefore, the equation becomes: \[pH = pKa + \log{\frac{[\mathrm{B}]}{[\mathrm{BH^{+}}]}}\] where, \(pKa = pKb - 14\) (since \(\mathrm{pKA + pKb = 14}\)) We know from the question that \(\mathrm{pKb = 9.0}\), so we can find \(\mathrm{pKa}\).

Calculate the pKa

Using the given pKb value, calculate the pKa: \[pKa = 14 - 9.0 = 5\] Now we can substitute pH, pKa, and \([\mathrm{B}]\) into the equation and solve for \([\mathrm{BH^{+}}]\).

Calculate the concentration of \(\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{NH}_{3}^{+}\)

Plug the values of pH, pKa, and \([\mathrm{B}]\) into the equation: \[4.20 = 5 + \log{\frac{0.50}{[\mathrm{BH^{+}}]}}\] Solve for \([\mathrm{BH^{+}}]\): \[4.20 - 5 = \log{\frac{0.50}{[\mathrm{BH^{+}}]}}\] \[-0.80 = \log{\frac{0.50}{[\mathrm{BH^{+}}]}}\] \[\frac{0.50}{[\mathrm{BH^{+}}]} = 10^{-0.80}\] Now find \([\mathrm{BH^{+}}]\): \[[\mathrm{BH^{+}}]=\frac{0.50}{10^{-0.80}}\] \[[\mathrm{BH^{+}}] \approx 0.158 M\] So, in this buffer solution, the concentration of \(\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{NH}_{3}^{+}\) is approximately \(0.158 M\). #b. Calculate the \(\mathrm{pH}\) after \(4.0 \mathrm{g} \mathrm{NaOH}(s)\) is added to \(1.0 \mathrm{L}\) of this solution.#

Calculate the moles of \(\mathrm{NaOH}\) added

We are given that 4.0 g of \(\mathrm{NaOH}\) is added to the solution. First, let's find the moles of \(\mathrm{NaOH}\): moles of \(\mathrm{NaOH} = \frac {mass}{molar\,mass}\) The molar mass of \(\mathrm{NaOH}\): Na = 22.99 g/mol, O = 16.00 g/mol, and H = 1.01 g/mol, hence \(22.99 + 16.00 + 1.01 \approx 40 g/mol\) \[\textrm{moles of }\mathrm{NaOH} = \frac {4.0 g}{40 \, g/mol} = 0.1\, mol\]

Reaction between the \(\mathrm{NaOH}\) and \(\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{NH}_{3}^{+}\)

When \(\mathrm{NaOH}\) is added to the solution, it reacts with the conjugate acid \(\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{NH}_{3}^{+}\) \[\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{NH}_{3}^{+} + \mathrm{OH^{-}} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5}\mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O}\] As \(\mathrm{0.1\,mol}\) of \(\mathrm{NaOH}\) is added, the \([\mathrm{BH^{+}}]\) will decrease by \(\mathrm{0.1\,mol}\), and the \([\mathrm{B}]\) will increase by \(\mathrm{0.1\,mol}\): \[[\mathrm{BH^{+}}] = 0.158 M - 0.1 M = 0.058 M\] \[[\mathrm{B}] = 0.50 M + 0.1 M = 0.60 M\]

Calculate the new pH

Now use the Henderson-Hasselbalch equation with the new concentrations: \[pH = pKa + \log{\frac{[\mathrm{B}]}{[\mathrm{BH^{+}}]}}\] \[pH = 5 + \log{\frac{0.60}{0.058}}\] \[pH \approx 5.78\] After adding 4.0 g of \(\mathrm{NaOH}\), the new pH of this solution is approximately 5.78.

Key concepts

Understanding the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is fundamental in biochemistry and pharmaceutical science, as it relates pH, a measure of acidity, to the concentrations of an acid and its conjugate base in a solution. It is presented in the form:
\[pH = pKa + \text{log}\left(\frac{[A^-]}{[HA]}\right)\]
where pH is the measure of the hydrogen ion concentration, pKa is the acid dissociation constant, which provides a measure of the strength of the acid in question, [A^-] is the molar concentration of the conjugate base, and [HA] is the molar concentration of the acid.

In practice, this equation is crucial for creating buffer solutions that can maintain a stable pH, an important aspect in many biological and chemical processes. To effectively use this equation, one should understand the concept of pKa and how pH relates to the ratio between the conjugate base and acid. When manipulating the components of the equation to solve for an unknown variable, logarithmic and exponential operations are often employed.

pH Calculation: An Essential Skill in Chemistry

The pH of a solution is a numeric scale used to specify the acidity or basicity (alkalinity) of an aqueous solution. It is roughly the negative logarithm of the hydrogen ion concentration, and calculated using the formula:
\[pH = -\text{log}[H^+]\]
For weaker acids and bases, which do not dissociate completely, pH calculation becomes less straightforward and relies on equilibrium constants and concentrations of the species in solution. Understanding and executing pH calculations are critical for students and researchers in the field, as pH can affect the behavior of molecules and the outcomes of reactions.

It is worth noting that pH is temperature-dependent, with a neutrality point at pH 7 only at 25°C. Outside of this temperature, the neutrality point changes. In industrial and lab settings, accurately calculating and adjusting pH is vital for process control and safety.

Weak Acid-Base Equilibria: A Delicate Balance

Weak acid-base equilibria involve reversible reactions where weak acids or bases do not dissociate completely in water, setting up an equilibrium between the undissociated and dissociated forms (conjugate pairs). The equilibrium is described by an equilibrium constant, Ka for weak acids and Kb for weak bases, which are measures of the strength of the acids and bases.

For example, \[HA \rightleftharpoons H^+ + A^-\] Calculating the concentrations of these species at equilibrium involves both the equilibrium constant and the initial concentrations. During calculations, assumptions such as the neglect of water's ionization or the approximation that the change in concentration of the reactants is negligible compared to the initial concentration can simplify the problem.

In buffer systems, these equilibria are essential, as they involve a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid), allowing the system to resist changes in pH when small amounts of acid or base are added.