Question

Calculate the \(\mathrm{pH}\) change when \(10.0 \mathrm{~mL}\) of \(0.100-\mathrm{M}\) \(\mathrm{NaOH}\) is added to \(90.0 \mathrm{~mL}\) pure water, and compare the \(\mathrm{pH}\) change with that when the same amount of \(\mathrm{NaOH}\) solution is added to \(90.0 \mathrm{~mL}\) of a buffer consisting of \(1.00-\mathrm{M} \mathrm{NH}_{3}\) and \(1.00-\mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\). Assume that the vol- umes are additive. \(K_{\mathrm{b}}\) of \(\mathrm{NH}_{3}=1.8 \times 10^{-5}\)

Short answer

The pH change is 5 in water and negligible in the buffer.

Step-by-step solution

Calculate moles of NaOH added

To find the moles of NaOH, use the formula \( ext{moles} = ext{concentration} imes ext{volume} \). Given that the concentration is \( 0.100 ext{ M} \) and the volume is \( 10.0 ext{ mL} = 0.010 ext{ L} \), the moles of NaOH added is \( 0.100 ext{ M} \times 0.010 ext{ L} = 0.001 ext{ moles} \).

Calculate pH of pure water after NaOH addition

Initially, pure water has a \( ext{pH} = 7 \). After adding NaOH, the resulting solution contains these additional moles of OH\(^{-} \). The total volume is \( 10.0 ext{ mL} + 90.0 ext{ mL} = 100.0 ext{ mL} \), or \( 0.100 ext{ L} \). The concentration of OH\(^{-} \) is \( \frac{0.001}{0.100} = 0.010 ext{ M} \). The \( ext{pOH} = - ext{log}(0.010) = 2 \). Therefore, \( ext{pH} = 14 - ext{pOH} = 12 \).

Calculate pH of buffer before NaOH addition

A buffer solution of NH\(_{3}\) and NH\(_{4}^{+}\) has pH calculated using the \( K_b \). Using the equation: \[ ext{pH} = 14 - ext{pOH} = 14 - (- ext{log}( ext{K}_b) + ext{log}(rac{[NH_{3}]}{[NH_4^+]})) \]. As \( [NH_{3}] = [NH_{4}^{+}] \), the solution is at equilibrium, so \( ext{pOH} = - ext{log}( ext{K}_b) = - ext{log}(1.8 \times 10^{-5}) \approx 4.74 \), and then \( ext{pH} = 14 - 4.74 = 9.26 \).

Calculate pH of buffer after NaOH addition

The addition of 0.001 moles of NaOH will react with NH\(_{4}^{+}\) to form more NH\(_{3}\). The concentrations after reaction are \( [NH_3] = 1.001 ext{ M} \) and \( [NH_4^+] = 0.999 ext{ M} \). Use the Henderson-Hasselbalch equation \( ext{pH} = ext{pK}_b + ext{log}(rac{[NH_3]}{[NH_4^{+}]}) \). Calculate \( ext{pK}_b = - ext{log}(1.8 \times 10^{-5}) \approx 4.74 \), then \( ext{pH} = 14 - (4.74 - ext{log}(rac{1.001}{0.999})) \approx 9.26 \) because the ratio is close to 1.

Compare pH changes

In pure water, the pH increases from 7 to 12, a change of 5 units. In the buffer, the pH changes negligibly, remaining approximately 9.26. Thus, the buffer resists pH change much more effectively than pure water.

Key concepts

pH Change

Understanding pH change is crucial when adding acids or bases to solutions. pH measures how acidic or basic a solution is, ranging from 0 (very acidic) to 14 (very basic), with 7 being neutral. When you add a small volume of a strong base, like NaOH, to water, it significantly increases the pH. This is because NaOH dissociates completely in water, increasing the concentration of hydroxide ions (OH\(^-\)).

The pH change implies how the balance of hydrogen and hydroxide ions shifts. For pure water, initial pH is 7. After adding NaOH, the increased concentration of OH\(^-\) raises the pH to 12.

Buffers, however, resist drastic pH changes. Adding the same NaOH to a buffer solution, like one containing NH\(_3\) and NH\(_4^+\), produces negligible change in pH. Exploring pH changes in these contexts enhances understanding of solution dynamics and behavior.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch Equation is valuable in calculating the pH of buffer solutions. This equation is derived from the equilibrium constants of acids and bases, providing a practical way to determine the buffer's pH. It is expressed as:
\[pH = pK_a + \log\left(\frac{[\text{base}]}{[\text{acid}]}\right)\]

For a buffer containing ammonia (NH\(_3\)) and its conjugate acid (NH\(_4^+\)), a similar equation for bases (using pK\(_b\)) is:
\[pH = 14 - (pK_b + \log\left(\frac{[\text{base}]}{[\text{acid}]}\right))\]

In this context, ammonia acts as the base, and ammonium ion acts as the acid. When calculating the pH after adding NaOH to the buffer, the Henderson-Hasselbalch equation simplifies the process. Since the concentrations of NH\(_3\) and NH\(_4^+\) don't vastly differ after NaOH addition, the pH remains stable, highlighting the buffer's effectiveness in resisting pH changes. Understanding this equation aids in predicting and controlling pH in various chemical and biological systems.

Acid-Base Equilibria

Acid-base equilibria underpin our understanding of chemical reactions involving acids and bases. These equilibria describe the balance between an acid (proton donor) and its conjugate base (proton acceptor), key in understanding how buffers work.

Ammonia (NH\(_3\)), a weak base, when combined with ammonium chloride (NH\(_4\)Cl), forms an effective buffer due to this equilibrium:
NH\(_3\) + H\(_2\)O \(\rightleftharpoons\) NH\(_4^+\) + OH\(^-\)

The equilibrium constant, known as K\(_b\), reflects ammonia's strength as a base. Adding a strong base like NaOH shifts this equilibrium, yet buffers like NH\(_3\)/NH\(_4^+\) resist significant pH change. This occurs because the excess OH\(^-\) reacts with the ammonium ions to form more ammonia, consuming the excess base and stabilizing pH.

A deeper understanding of acid-base equilibria allows insight into not only buffer systems but also broader applications across chemistry, biology, and environmental science. Recognizing these interactions underpins the design of systems that require precise pH control.