Question

A \(10.0-\mathrm{mL}\) solution of \(0.300 \mathrm{M} \mathrm{NH}_{3}\) is titrated with a \(0.100 \mathrm{M} \mathrm{HCl}\) solution. Calculate the \(\mathrm{pH}\) after the following additions of the \(\mathrm{HCl}\) solution: (a) \(0.0 \mathrm{~mL}\) (b) \(10.0 \mathrm{~mL}\) (c) \(20.0 \mathrm{~mL}\) (d) \(30.0 \mathrm{~mL}\) (e) \(40.0 \mathrm{~mL}\)

Short answer

(a) pH = 11.37, (b) pH = 9.56, (c) pH = 8.96, (d) pH = 5.19, (e) pH = 1.70

Step-by-step solution

Initial Condition (0.0 mL HCl)

Before any hydrochloric acid is added, the solution contains only ammonia, which is a weak base. The initial concentration of ammonia, \( \text{NH}_3 \), is \( 0.300 \text{ M} \). To calculate the initial pH, we first need to calculate the concentration of hydroxide ions \( \text{OH}^- \) in the solution, using the base dissociation constant \( K_b \) for ammonia. This is given by:\[ K_b = 1.8 \times 10^{-5} \]For the reaction \( \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \), we use the equation:\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]Assuming that \( x \) is the concentration of \( \text{OH}^- \) formed, we have:\[ x^2 = K_b \times (0.300 - x) \approx K_b \times 0.300 \]Solving for \( x \), we find:\[ x = \sqrt{1.8 \times 10^{-5} \times 0.300} \approx 2.34 \times 10^{-3} \text{ M} \]Now, calculate the pOH:\[ \text{pOH} = -\log(2.34 \times 10^{-3}) \approx 2.63 \]The pH is:\[ \text{pH} = 14 - \text{pOH} = 11.37 \]

Addition of 10.0 mL HCl

When 10.0 mL of \( 0.100 \text{ M} \) HCl is added, it reacts completely with \( \text{NH}_3 \) to form \( \text{NH}_4^+ \). The moles of \( \text{HCl} \) added are:\[ 0.010 \text{ L} \times 0.100 \text{ M} = 0.001 \text{ moles} \]The initial moles of \( \text{NH}_3 \) are:\[ 0.010 \text{ L} \times 0.300 \text{ M} = 0.003 \text{ moles} \]After addition, the remaining moles of \( \text{NH}_3 \) are:\[ 0.003 - 0.001 = 0.002 \text{ moles} \]The volume of the solution is now \( 20.0 \text{ mL} \), or \( 0.020 \text{ L} \). The concentration of \( \text{NH}_3 \) is:\[ \frac{0.002}{0.020} = 0.100 \text{ M} \]The concentration of \( \text{NH}_4^+ \) (which is equal to the moles of \( \text{HCl} \) added) is also \( 0.050 \text{ M} \) because:\[ \frac{0.001}{0.020} = 0.050 \text{ M} \]We use the Henderson-Hasselbalch equation to find the pH:\[ \text{pH} = pK_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]Here, \( pK_a = 14 - pK_b \approx 14 - 4.74 = 9.26 \) (for \( \text{NH}_4^+ \))Thus:\[ \text{pH} = 9.26 + \log\left(\frac{0.100}{0.050}\right) = 9.26 + 0.301 = 9.561 \]

Addition of 20.0 mL HCl

After adding 20.0 mL of \( 0.100 \text{ M} \) HCl, we calculate the moles of \( \text{NH}_3 \) and \( \text{HCl} \) used:- Moles of \( \text{HCl} \) added: \( 0.020 \text{ L} \times 0.100 \text{ M} = 0.002 \text{ moles} \)- Initial moles of \( \text{NH}_3 \) were 0.003. After reacting with 0.002 moles of \( \text{HCl} \), 0.001 moles of \( \text{NH}_3 \) remainThe total volume is \( 30.0 \text{ mL} \) or \( 0.030 \text{ L} \):- Concentration of \( \text{NH}_3 \): \( \frac{0.001}{0.030} = 0.0333 \text{ M} \)- Concentration of \( \text{NH}_4^+ \): \( \frac{0.002}{0.030} = 0.0667 \text{ M} \)Using Henderson-Hasselbalch:\[ \text{pH} = 9.26 + \log\left(\frac{0.0333}{0.0667}\right) = 9.26 - 0.301 = 8.959 \]

Addition of 30.0 mL HCl

Upon adding 30.0 mL of \( 0.100 \text{ M} \) HCl, we have:- Moles of \( \text{HCl} \) added: \( 0.030 \text{ L} \times 0.100 \text{ M} = 0.003 \text{ moles} \)- Initial moles of \( \text{NH}_3 \) = 0.003This completely reacts with the \( \text{HCl} \), so all \( \text{NH}_3 \) becomes \( \text{NH}_4^+ \).The total concentration of \( \text{NH}_4^+ \) in the solution (40.0 mL or 0.040 L) is:\[ \frac{0.003}{0.040} = 0.075 \text{ M} \]Since \( \text{NH}_4^+ \) is a weak acid, the pH can be found using \( K_a \) of \( \text{NH}_4^+ \):\[ K_b = 1.8 \times 10^{-5} \rightarrow K_a = \frac{1 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10} \]Assuming \( y \) is the concentration of \( \text{H}^+ \) ions:\[ y^2 = K_a \times 0.075 \rightarrow y = \sqrt{5.56 \times 10^{-10} \times 0.075} \approx 6.45 \times 10^{-6} \text{ M} \]\[ \text{pH} = -\log(6.45 \times 10^{-6}) \approx 5.19 \]

Addition of 40.0 mL HCl

Adding 40.0 mL of \( 0.100 \text{ M} \) HCl results in:- Moles of \( \text{HCl} \): \( 0.040 \text{ L} \times 0.100 \text{ M} = 0.004 \text{ moles} \)After reacting with 0.003 moles of \( \text{NH}_3 \), 0.001 moles of \( \mathrm{HCl} \) remain unreacted.- The unreacted \( \text{HCl} \) in 50 mL or 0.050 L:\[ \frac{0.001}{0.050} = 0.020 \text{ M} \]Since this is a strong acid, the pH:\[ \text{pH} = -\log(0.020) \approx 1.70 \]

Key concepts

pH calculation

Calculating the pH of a solution is crucial in understanding its acidity or basicity. Let's explore some key concepts crucial for these calculations. The pH scale is logarithmic and ranges from 0 to 14, which measures the concentration of hydrogen ions (\(\text{H}^+\)) in a solution. A lower pH value indicates higher acidity, while a higher pH indicates basicity. The formula is given by:\[\text{pH} = -\log[\text{H}^+]\]For solutions with a weak base, like ammonia, you often first calculate the pOH and then use the relation between pH and pOH:\[\text{pH} + \text{pOH} = 14\]This calculation becomes especially relevant when performing titrations, where accurate pH values help you understand the progression of the reaction between acid and base.

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a powerful tool used to calculate the pH of a solution containing a weak acid and its conjugate base, or a weak base and its conjugate acid. This equation is beneficial during titrations involving weak bases like ammonia:\[\text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]\right)\]For ammonia, during a titration with hydrochloric acid (HCl), this equation helps calculate the pH after partial neutralization, where both ammonia (\(\text{NH}_3\)) and its conjugate acid ammonium (\(\text{NH}_4^+\)) are present in the solution. The Henderson-Hasselbalch equation simplifies the process of determining the pH at different stages of a titration, reflecting how the ratio of base to acid affects the solution's pH directly.

acid-base reactions

Acid-base reactions entail the mutual exchange of protons (\(\text{H}^+\)) between the acid and the base. During a titration of ammonia (\(\text{NH}_3\)), a weak base, with hydrochloric acid (\(\text{HCl}\)), a strong acid, the reaction will look like this:\[\text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4^+ + \text{Cl}^-\]

• The acid (\(\text{HCl}\)) donates a proton, turning ammonia (\(\text{NH}_3\)) into ammonium (\(\text{NH}_4^+\)).
• This reaction affects the pH as it progresses, highlighting the importance of tracking the amount of acid added.
• Understanding these reactions assists in calculating how the concentrations of the substances change, which is crucial for determining the solution's pH at various points of titration.

ammonia titration

Ammonia titration involves the gradual addition of a titrant—usually a strong acid like hydrochloric acid—to a solution of ammonia. The objective is to determine the ammonia's concentration or how the chemical reaction progresses.

• Ammonia, being a weak base, reacts with the acid producing its conjugate acid, ammonium (\(\text{NH}_4^+\)).
• At different stages of this titration, the pH changes significantly, allowing students to track these changes and improve their understanding of acid-base equilibrium.
• Using concepts like the Henderson-Hasselbalch equation can simplify calculations, predicting the pH at various stages and effectively understanding the interaction between acid and base components.

This process teaches students not only about finding the endpoint of the titration but also about the chemical principles of buffer solutions and the broader context of acid-base chemistry.