Question

For each solution, calculate the initial and final pH after the addition of 0.010 mol of \(\mathrm{NaOH}\) a. 250.0 \(\mathrm{mL}\) of pure water b. 250.0 \(\mathrm{mL}\) of a buffer solution that is 0.195 \(\mathrm{M}\) in \(\mathrm{HCHO}_{2}\) and 0.275 \(\mathrm{M}\) in \(\mathrm{KCHO}_{2}\) c. 250.0 \(\mathrm{mL}\) of a buffer solution that is 0.255 \(\mathrm{M}\) in \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\) and 0.235 \(\mathrm{M}\) in \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3} \mathrm{Cl}\)

Short answer

Part A: Initial pH = 7, final pH calculated from pOH of added NaOH. Part B: Initial pH from Henderson-Hasselbalch equation, final pH adjusted for changes in buffer concentration post-NaOH addition. Part C: Similar approach to Part B with different buffer components and pKa.

Step-by-step solution

Part A: Initial pH of pure water

The pH of pure water is 7 since it is neutral. This is because at 25 degrees Celsius, the concentration of hydrogen ions \( [H^+] \) and hydroxide ions \( [OH^-] \) in pure water are both \(1 \times 10^{-7} \mathrm{M}\), so the pH is calculated as \(-\log [H^+] = -\log (1 \times 10^{-7}) = 7\).

Part A: Final pH after adding NaOH to water

The number of moles of \(\mathrm{NaOH}\) added is 0.010 moles. When this fully dissociates in 250 mL of water, the concentration of \(OH^−\) ions becomes \(0.010 \mathrm{moles} / 0.250 \mathrm{L} = 0.040 \mathrm{M}\). To find the pOH, use the formula \(pOH = -\log [OH^-]\), which gives \(pOH = -\log (0.040)\). The pH is then calculated using \(pH = 14 - pOH\).

Part B: Initial pH of buffer solution

Henderson-Hasselbalch equation is used to calculate the initial pH of the buffer solution: \(pH = pKa + \log \left(\frac{[A^-]}{[HA]}\right)\). The pKa of \(\mathrm{HCHO}_2\) needs to be known; assuming it to be 3.74, the initial pH is calculated as \(pH = pKa + \log \left(\frac{[\mathrm{KCHO}_2]}{[\mathrm{HCHO}_2]}\right) = pKa + \log \left(\frac{0.275}{0.195}\right)\).

Part B: Final pH after adding NaOH to buffer solution

The buffer solution will resist changes in pH. After adding 0.010 moles of \(\mathrm{NaOH}\), it will react with the \(\mathrm{HCHO}_2\) to form \(\mathrm{KCHO}_2\). The final concentrations of \(\mathrm{HCHO}_2\) and \(\mathrm{KCHO}_2\) should be recalculated, and the new pH can be calculated again using the Henderson-Hasselbalch equation.

Part C: Initial pH of buffer solution

Again, we use the Henderson-Hasselbalch equation: \(pH = pKa + \log \left(\frac{[base]}{[acid]}\right)\). The pKa of \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{NH}_3^+\) needs to be known; assuming it to be 10.8, the initial pH is calculated as \(pH = pKa + \log \left(\frac{[\mathrm{CH}_3\mathrm{CH}_2\mathrm{NH}_2]}{[\mathrm{CH}_3\mathrm{CH}_2\mathrm{NH}_3\mathrm{Cl}]}\right)\).

Part C: Final pH after adding NaOH to buffer solution

When 0.010 moles of \(\mathrm{NaOH}\) are added to the buffer, it will react with \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{NH}_3^+\) to form \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{NH}_2\) and water. The concentrations should be updated, and the final pH calculated with the updated ratio using the Henderson-Hasselbalch equation.

Key concepts

Understanding Buffer Solutions

A buffer solution is like a chemical bodyguard for pH levels; it resists changes in pH when acids or bases are added to it. Imagine pouring a little bit of lemon juice into a cup of water; the water's pH would change quickly. But if you had a buffer solution instead of just water, it would fight back, keeping the pH almost the same.
Buffers work because they contain a weak acid and its conjugate base, or a weak base and its conjugate acid. Each part takes a turn protecting the pH: when an acid is added, the base 'grabs' those extra H+ ions, and when a base is added, the acid 'hands over' some H+ ions.

Why do we care?

Because life itself loves steadiness. Our bodies, for one, are natural examples of buffered systems, keeping our blood pH tightly controlled. In the lab, buffers help maintain ideal conditions for reactions. Without buffers, biochemistry would be a wild rollercoaster of pH swings.

The Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is like a GPS for navigating pH levels in a buffer solution. It tells you exactly where you're at and helps predict where you'll go when acids or bases join the party.
Here's the formula to remember: \(pH = pKa + \[log\left(\frac{[A^-]}{[HA]}\right)\]\), where \(pKa\) is the acid dissociation constant, \([A^-]\) is the concentration of the conjugate base, and \([HA]\) is the concentration of the acid.

How to use it

First, you need to know the \(pKa\), which tells you how strong the acid in your buffer is. The equation then uses the ratio of the base to the acid to find the pH. When the concentrations are equal, the pH equals the \(pKa\), and when the base is higher, the pH goes up, and vice versa.
The real beauty is when you add acid or base to your buffer; the equation helps recalculate the new pH, showing the buffer's power to minimize change.

Acid-base Titration

Acid-base titration is a scientific detective game, where you're figuring out the unknown concentration of an acid or a base. You do this by slowly adding a reactant of known concentration until there's just the right amount to neutralize the unknown reactant—this magic moment is called the equivalence point.
To spot this milestone, you use indicators that change color at certain pH levels, or you measure the pH directly as you add the titrant. The volume of titrant used at the equivalence point tells you how much unknown reactant you had.

Real-life titration

Imagine a chef tasting soup and adding just enough salt until it's perfect—not too bland, not too salty. That's titration, kitchen-style! Scientists use titration in labs for tasks like purifying water or finding the quality of wine. This process is crucial for getting just-right reactions and making sure that medicines are safe and effective.