Question

(a) Calculate the pH of a buffer that is \(0.105 \mathrm{M}\) in \(\mathrm{NaHCO}_{3}\) and \(0.125 \mathrm{M}\) in \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). (b) Calculate the \(\mathrm{pH}\) of a solution formed by mixing \(65 \mathrm{~mL}\) of \(0.20 \mathrm{M} \mathrm{NaHCO}_{3}\) with \(75 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\)

Short answer

For part (a), the pH of the buffer solution is approximately \(10.50\). For part (b), after mixing the given volumes and concentrations of NaHCO3 and Na2CO3, the pH of the solution is approximately \(10.19\).

Step-by-step solution

Determine the pKa value

As these solutions contain a weak acid (HCO3-) and its conjugate base (CO3^2-), we need to determine the pKa value for the given reaction. This reaction is given by: \( HCO_{3}^{-} \rightleftharpoons H^{+} + CO_{3}^{2-} \) For HCO3- / CO3^2-, the pKa value is 10.33. We will use this value in the following calculations.

Calculate the pH for part (a)

Now, use the Henderson-Hasselbalch equation to calculate the pH of part (a): \( \mathrm{pH} = \mathrm{pKa} + \log \frac{[\mathrm{Base}]}{[\mathrm{Acid}]} \) where: - Base: CO3^2- ion (from Na2CO3) - Acid: HCO3- ion (from NaHCO3) Insert the given values: \( \mathrm{pH} = 10.33 + \log \frac{0.125\;\mathrm{M}}{0.105\;\mathrm{M}} \) Calculate the pH: \( \mathrm{pH} = 10.33 + 0.17 \approx 10.50 \) So, the pH of the buffer solution in part (a) is approximately 10.50.

Calculate the moles of each component in part (b)

Before calculating the pH for part (b), we need to find the moles of NaHCO3 and Na2CO3 in the given volumes: Moles of NaHCO3: \( \mathrm{moles\; NaHCO_{3}} = \mathrm{volume} \times \mathrm{concentration} \) \( \mathrm{moles\; NaHCO_{3}} = 65\;\mathrm{mL} \times 0.20\;\mathrm{M} = 13\;\mathrm{mmol} \) Moles of Na2CO3: \( \mathrm{moles\; Na_{2}CO_{3}} = \mathrm{volume} \times \mathrm{concentration} \) \( \mathrm{moles\; Na_{2}CO_{3}} = 75\;\mathrm{mL} \times 0.15\;\mathrm{M} = 11.25\;\mathrm{mmol} \)

Calculate the concentrations of each component after mixing

Now find the new concentrations of NaHCO3 and Na2CO3 in the mixed solution: Total volume: \(65\;\mathrm{mL} + 75\;\mathrm{mL} = 140\;\mathrm{mL}\) New concentration of NaHCO3: \( \mathrm{[NaHCO_{3}]} = \frac{13\;\mathrm{mmol}}{140\;\mathrm{mL}} = 0.093\;\mathrm{M} \) New concentration of Na2CO3: \( \mathrm{[Na_{2}CO_{3}]} = \frac{11.25\;\mathrm{mmol}}{140\;\mathrm{mL}} = 0.080\;\mathrm{M} \)

Calculate the pH for part (b)

Finally, we'll use the new concentrations in the Henderson-Hasselbalch equation to calculate the pH of part (b): \( \mathrm{pH} = 10.33 + \log \frac{0.080\;\mathrm{M}}{0.093\;\mathrm{M}} \) Calculate the pH: \( \mathrm{pH} = 10.33 - 0.14 \approx 10.19 \) So, the pH of the solution formed by mixing the two components in part (b) is approximately 10.19.

Key concepts

Henderson-Hasselbalch Equation

Understanding the Henderson-Hasselbalch equation is integral to predicting the pH level of buffer solutions. This equation represents a direct relationship between the pH of a solution and the concentrations of a weak acid and its conjugate base.

In its simplest form, the equation reads: \[\begin{equation}\mathrm{pH} = \mathrm{pKa} + \log \frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\end{equation}\]Here, \(\mathrm{pKa}\) is the negative logarithm of the acid dissociation constant (\(Ka\)) for the weak acid in the buffer. The \(\log\) refers to the logarithm base 10, and \([\mathrm{Base}]\) and \([\mathrm{Acid}]\) are the molar concentrations of the conjugate base and the weak acid, respectively. The equation thus establishes a method to calculate the pH of a solution when you know the concentration ratio of the buffer components. To make the concept even more accessible:

• \(\mathrm{pKa}\) is like a 'pivot point' around which the pH can be adjusted.
• \([\mathrm{Base}]\)/\([\mathrm{Acid}]\) ratio indicates how much the buffer can resist changes in pH.
• If \([\mathrm{Base}]\) equals \([\mathrm{Acid}]\), the log term will be zero, making the pH equal to the \(\mathrm{pKa}\).

The Henderson-Hasselbalch equation is used in the textbook exercise to solve both (a) and (b) parts, providing a clear example of its practical application in buffer pH calculation.

Conjugate Acid-Base Pairs

To dive deeper into what makes buffers work, we examine conjugate acid-base pairs. These pairs consist of a weak acid and its corresponding conjugate base (or a weak base and its conjugate acid). They're crucial in buffering solutions because they neutralize excess acids or bases that might otherwise significantly change the pH.

In a chemical reaction, conjugate acid-base pairs are related by the loss or gain of a proton (\(H^{+}\)). For example, bicarbonate (\(HCO_{3}^{-}\)) and carbonate (\(CO_{3}^{2-}\)) form such a pair. With the addition of a proton, \(HCO_{3}^{-}\) can become carbonic acid (\(H_{2}CO_{3}\)), while the loss of a proton transforms it into \(CO_{3}^{2-}\). The textbook example you’re studying uses this exact pair in the buffer system. Key pointers include:

• Conjugate pairs maintain the pH by reacting with added acids or bases.
• The stronger the acid, the weaker its conjugate base, and vice versa.
• Understanding these pairs helps predict buffer behavior and solve related problems.

Stoichiometry

Stoichiometry is the aspect of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It helps us understand the proportions and amounts of substances that interact.

In the context of our buffer problem, stoichiometry allows for the calculation of the moles of each component in the buffer system. This is critical since knowing the amount of an acid and its conjugate base is a prerequisite for using the Henderson-Hasselbalch equation. To shed more light on the process:

• Moles are a measure for counting particles, similar to how a dozen counts eggs.
• The students calculated the moles of \(NaHCO_{3}\) and \(Na_{2}CO_{3}\) using their respective volumes and concentrations from part (b).
• This step is essential to later finding their molar concentrations in the final solution after mixing.

The conversion from volumes and molarities to moles is a fundamental stoichiometric step, and it helps in mastering chemistry problem-solving.

Bicarbonate Buffer System

The bicarbonate buffer system is one of the body's most important mechanisms for maintaining pH homeostasis, particularly in blood plasma. In our problem, we use the bicarbonate (\(HCO_{3}^{-}\)) and carbonate (\(CO_{3}^{2-}\)) system as an example of a buffer solution. This system showcases how equilibria work to keep the pH relatively stable, despite additions of acids or bases.

Here’s a closer look at this system:

• The bicarbonate ion acts as a weak acid that can react with any added bases.
• The carbonate ion functions as the conjugate base to the bicarbonate ion, ready to neutralize any added acids.
• This dual action helps buffers resist drastic changes in pH.

The textbook exercise leverages this buffer system to illustrate how pH remains relatively unchanged even when the proportions of bicarbonate and carbonate are altered after mixing different volumes of their sodium salt solutions.