Question

Tris [tris(hydroxymethyl)aminomethane] is a common buffer for studying biological systems: (a) Calculate the \(\mathrm{pH}\) of the tris buffer after mixing \(15.0 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) \(\mathrm{HCl}\) solution with \(25.0 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) tris. (b) This buffer was used to study an enzyme-catalyzed reaction. As a result of the reaction, 0.00015 mole of \(\mathrm{H}^{+}\) was consumed. What is the \(\mathrm{pH}\) of the buffer at the end of the reaction? (c) What would be the final \(\mathrm{pH}\) if no buffer were present?

Short answer

(a) 7.90; (b) 8.01; (c) 2.43

Step-by-step solution

Initial Moles Calculation

To find the pH of the buffer, first calculate the initial moles of HCl and tris. For HCl, the moles are given by \( n = C \times V = 0.10 \, \text{M} \times 0.015 \, \text{L} = 0.0015 \, \text{mol} \). For tris, the moles are \( n = 0.10 \, \text{M} \times 0.025 \, \text{L} = 0.0025 \, \text{mol} \).

Reaction Completion

The strong acid HCl will react completely with tris, a weak base. The reaction is \( \text{tris} + \text{HCl} \rightarrow \text{tris-H}^{+} + \text{Cl}^{-} \). After the reaction, \(0.0015\) moles of tris have reacted, leaving \(0.0010\) moles of unreacted tris and \(0.0015\) moles of tris-H\(^+\).

Calculate pH Using Henderson-Hasselbalch Equation

Use the Henderson-Hasselbalch equation: \[ \text{pH} = \text{pK}_\text{a} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]. The \( \text{pK}_\text{a} \) for tris is 8.08. Thus, \[ \text{pH} = 8.08 + \log \left( \frac{0.0010}{0.0015} \right) \]. \( \log(0.6667) \approx -0.1761 \), so \( \text{pH} \approx 8.08 - 0.1761 = 7.90 \).

Adjust pH After Reaction Consumption

A consumption of 0.00015 moles of \(\text{H}^+\) means an increase of 0.00015 moles in the unprotonated form of tris. Now, there are \( 0.00115 \) moles of unreacted tris and \( 0.00135 \) moles of tris-H\(^+\). Use the Henderson-Hasselbalch equation again: \[ \text{pH} = 8.08 + \log \left( \frac{0.00115}{0.00135} \right) \]. Since \( \log(0.8519) \approx -0.0698 \), \( \text{pH} \approx 8.08 - 0.0698 = 8.01 \).

Final pH Without Buffer

If no buffer was present, the final solution would mainly consist of water. The remaining \(0.00015\) moles of \(\text{H}^+\) ions in \(0.040 \text{ L}\) \(25 \text{ mL} + 15 \text{ mL}\) gives a concentration of \(\frac{0.00015}{0.040} = 0.00375 \text{ M}\). Thus, \(\text{pH} = -\log(0.00375) \approx 2.43 \).

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a critical tool in buffer solution chemistry. It provides a straightforward way to calculate the pH of buffer solutions, which are essential for maintaining stable pH levels in various chemical environments. This equation is written as:\[ \text{pH} = \text{pK}_\text{a} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]Here, \([\text{A}^-]\) represents the concentration of the base form of the buffer, and \([\text{HA}]\) is the concentration of the acid form. The term \(\text{pK}_\text{a}\) is the negative logarithm of the acid dissociation constant, which characterizes the strength of the acid in solution.

• The equation shows how your buffer solution will resist changes in pH upon the addition of small amounts of acids or bases.
• Using the equation involves substituting real-world concentrations to predict how your buffer will behave in different conditions.

For this exercise, understanding the Henderson-Hasselbalch equation is crucial because it allows us to predict and verify how much the pH of the tris buffer changes after HCl is added and after some \(\text{H}^+\) is consumed.

pH calculation

Calculating the pH in buffer solutions is fundamental in chemistry, especially when dealing with biological systems where pH levels need to be stable. The whole process hinges on understanding how the presence of a weak acid and its conjugate base counteract changes in hydrogen ion concentration.

• The pH calculation involves determining the concentrations of the acid and base forms of the buffer after the reaction.
• In this exercise, the initial pH calculation uses the Henderson-Hasselbalch equation once the initial amounts of tris and HCl have reacted.

For the pH after consumption of \(0.00015\) moles of \(\text{H}^+\), this calculation reflects shifts in the buffer's balance as more \(\text{H}^+\) ions are removed, highlighting the buffer's role in stabilizing pH.

Acid-base reactions

Acid-base reactions are central to understanding buffer solutions. In the context of this exercise, we examine the reaction between HCl, a strong acid, and tris, a weak base. The complete reaction can be expressed as:\[ \text{tris} + \text{HCl} \rightarrow \text{tris-H}^+ + \text{Cl}^- \]Here, the strong acid HCl dissociates completely in the solution, providing protons \(\text{H}^+\) that react with tris. This reaction reduces the amount of tris available while forming tris-H\(^+\), the conjugate acid of tris.

• The completion of this reaction is essential, as it affects the concentrations of the components needed for the pH calculations.
• Understanding how this acid-base interaction alters the chemistry of the solution helps us appreciate the stability buffer solutions provide by neutralizing unwanted pH changes.

This exercise illustrates how such reactions underpin buffer systems, allowing them to maintain steady pH, which is critical in living organisms and many industrial processes.

Moles and concentrations

In any chemical reaction, including those involving buffer solutions, understanding moles and concentrations is crucial. Moles tell us how much of a substance is present, while concentration (moles per unit volume) indicates how much of that substance is dispersed in the solution. In this exercise:

• The moles of each reactant (HCl and tris) are calculated by multiplying their concentration by the volume of solution, using the formula \( n = C \times V \).
• For HCl, 0.10 M concentration and 15.0 mL volume give 0.0015 moles, while for tris, 0.10 M concentration and 25.0 mL volume account for 0.0025 moles.

Understanding these initial moles allows us to predict how much of each reactant will remain after the acid-base reaction. This information is pivotal when applying the Henderson-Hasselbalch equation, as it directly influences the pH calculation and informs us about the buffer's efficacy in maintaining the solution’s pH even after changes from reactions. Proper mastery of moles and concentrations is indispensable in chemistry to ensure precise formulation and outcome of experiments.