Problem 46 - Chemistry The Central Science In Si Units

Question

Consider the titration of \(30.0 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{NH}_{3}\) with \(0.025 \mathrm{MHCl}\). Calculate the \(\mathrm{pH}\) after the following volumes of titrant have been added: (a) \(0 \mathrm{~mL},\) (b) \(20.0 \mathrm{~mL},\) (c) 59.0 \(\mathrm{mL}_{,}\) (d) \(60.0 \mathrm{~mL},\) (e) \(61.0 \mathrm{~mL},\) (f) \(65.0 \mathrm{~mL}\)

Step-by-Step Solution

Answer

(a) pH = 10.97, (b) pH = 9.55, (c) pH = 5.75, (d) pH = 4.93, (e) pH = 3.56, (f) pH = 2.88

Step 1: Identify initial conditions

Initially, we have 30 mL of 0.050 M NH3. NH3 is a weak base and partially dissociates in water, but initially, there's no HCl added, so the pH is determined by NH3 alone.

Step 2: Calculate initial pH

Use the base dissociation expression for NH3:NH3 + H2O ⇌ NH4+ + OH−The equilibrium expression is:\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} = 1.8 \times 10^{-5} \]Let the initial concentration of NH4+ and OH− be x. Therefore, the equation becomes:\[ 1.8 \times 10^{-5} = \frac{x^2}{0.050 - x} \]Since x is small, approximate: \[ 1.8 \times 10^{-5} = \frac{x^2}{0.050} \]Solving for x gives the [OH−], then calculate pOH and pH:\[ x \approx \sqrt{1.8 \times 10^{-5} \times 0.050} = 0.000943 \]pOH = -log(0.000943) = 3.03pH = 14 - pOH = 10.97

Step 3: Determine pH after 20 mL HCl added

Calculate the moles of NH3 and HCl:- Moles of NH3: \( 30 \times 0.050 / 1000 = 0.0015 \text{ moles} \)- Moles of HCl added: \( 20 \times 0.025 / 1000 = 0.0005 \text{ moles} \)This partially neutralizes NH3. Remaining NH3: 0.0015 - 0.0005 = 0.0010 moles.Calculate the pH using the Henderson-Hasselbalch equation:\[ \text{pH} = \text{pKa} + \log \left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]\[ \text{pKa} = 14 - 4.75 = 9.25 \] (for NH4+)Concentrations in the total volume (50 mL):\[ [\text{NH}_4^+] = \frac{0.0005}{0.050} = 0.010, \quad [\text{NH}_3] = \frac{0.0010}{0.050} = 0.020 \]\[ \text{pH} = 9.25 + \log \left(\frac{0.020}{0.010}\right) = 9.55 \]

Step 4: Calculate pH after 59 mL HCl added

A similar calculation for 59 mL of added acid will leave some NH3:- Moles of HCl: \( 59 \times 0.025 / 1000 = 0.001475 \text{ moles} \)Remaining NH3: 0.0015 - 0.001475 = 0.000025 moles.Using the buffer equation because of a large excess of NH4+:\[ \text{pH} = 9.25 + \log \left(\frac{0.000025}{0.001475}\right) \approx 5.75 \]

Step 5: Equivalence point calculation for 60 mL HCl

Equivalence point reached at 60 mL when all NH3 has reacted:- Moles of HCl: \( 60 \times 0.025 / 1000 = 0.0015 \text{ moles} \)At equivalence, the solution contains NH4+ (weak acid):\[ \text{NH}_4^+ \rightarrow \text{NH}_3 + \text{H}^+ \]\[ K_a = \frac{K_w}{K_b} = \frac{1 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10} \]Use the K_a expression for NH4+:\[ 5.56 \times 10^{-10} = \frac{x^2}{0.025} \]\[ x = \sqrt{5.56 \times 10^{-10} \times 0.025} \approx 1.18 \times 10^{-5} \]pH = -log(1.18 \times 10^{-5}) = 4.93

Step 6: Beyond equivalence after 61 mL HCl added

Adding 61 mL results in excess H+- Additional moles of HCl: \( 0.000025 \text{ moles of HCl} \)Total volume: 91 mL, H+ concentration:\[ [\text{H}^+] = \frac{0.000025}{0.091} \approx 2.75 \times 10^{-4} \]pH = -log(2.75 \times 10^{-4}) = 3.56

Step 7: Larger excess after 65 mL HCl added

Calculating for 65 mL of HCl,- Extra moles of HCl: \( (65-60) \times 0.025 / 1000 = 0.000125 \text{ moles} \)Total volume: 95 mL, H+ concentration:\[ [\text{H}^+] = \frac{0.000125}{0.095} \approx 1.32 \times 10^{-3} \]pH = -log(1.32 \times 10^{-3}) = 2.88

Key Concepts

Understanding Weak Bases

A weak base is a substance that can accept protons but does not completely disassociate in water. In this exercise, ammonium (NH_3) serves as a classic example of a weak base. Unlike strong bases, which dissociate completely, weak bases only partially split into their ions in solution.

When NH_3 is dissolved in water, it sets up an equilibrium where some of the base captures hydrogen ions from water to form ammonium ions (NH_4^+) and hydroxide ions (OH^−). This balance between the undissociated molecules and ions is an important feature of weak bases that affects the pH of the solution.

\[ ext{NH}_3(aq) + ext{H}_2O(l) \rightleftharpoons ext{NH}_4^+(aq) + ext{OH}^-(aq) \]

• They establish an equilibrium state rather than a full dissociation.
• The equilibrium expression is represented by the base dissociation constant, K_b.
• The K_b for ammonia is 1.8 \times 10^{-5}, indicating limited ionization.

Understanding the degree of this ionization is vital in determining the pH of a solution containing a weak base.

The Process of pH Calculation

The pH scale is an important measure in chemistry that helps determine the acidity or basicity of a solution. pH is calculated as the negative logarithm of the hydrogen ion concentration: \( ext{pH} = - ext{log}_{10}[ ext{H}^+] \).

When dealing with weak bases like NH_3, it is also crucial to determine the concentration of hydroxide ions (OH^−), as they will affect the solution's pH as well. This is often accomplished by first calculating the pOH and then finding pH using the relationship between pH and pOH:

\[ ext{pH} + ext{pOH} = 14 \]

• Use the equilibrium expression and value of K_b to find [ ext{OH}^-].
• Calculate  ext{pOH} using \(- ext{log}[ ext{OH}^-]\).
• Subsequently, find the  ext{pH} with the relationship:  ext{pH} = 14 - ext{pOH}.

This method forms the basis of pH calculation in many scenarios involving weak bases.

Exploring the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a powerful tool for estimating the pH of a solution during a titration involving weak acids or bases. Specifically, it provides a way to calculate pH based on the concentrations of the base and its conjugate acid.

The formula is:
\[ ext{pH} = ext{pK}_a + ext{log} \left(\frac{[ ext{Base}]}{[ ext{Acid}]}\right) \]. In this exercise, we converted K_b to K_a using the relation \( ext{pK}_a = 14 - ext{pK}_b \). With NH_3 being the weak base, and NH_4^+, its conjugate acid:

• At equilibrium, when some amount of titrant like HCl is added, [ ext{Base}] and [ ext{Acid}] must be determined.
• Calculate  ext{pH} using the known  ext{pK}_a and relative concentrations of NH_3 and NH_4^+.
• This equation is especially useful when dealing with buffer solutions, where pH remains relatively stable despite small additions of acid or base.

It simplifies the computation of pH by focusing on the ratio of concentrations rather than complex equilibrium expressions.