Problem 26 - Chemistry: Principles and Reactions

Question

WEB To make a buffer with \(\mathrm{pH}=3.0\) from \(\mathrm{HCHO}_{2}\) and \(\mathrm{CHO}_{2}^{-}\), (a) what must the [HCHO \(\left._{2}\right] /\left[\mathrm{CHO}_{2}^{-}\right]\) ratio be? (b) how many moles of \(\mathrm{HCHO}_{2}\) must be added to a liter of \(0.139 \mathrm{M}\) \(\mathrm{NaCHO}_{2}\) to give this \(\mathrm{pH}\) ? (c) how many grams of \(\mathrm{NaCHO}_{2}\) must be added to \(350.0 \mathrm{~mL}\) of \(0.159 \mathrm{MHCHO}_{2}\) to give this \(\mathrm{pH}\) ? (d) What volume of \(0.236 \mathrm{M} \mathrm{HCHO}_{2}\) must be added to \(1.00 \mathrm{~L}\) of a \(0.500 \mathrm{M}\) solution of \(\mathrm{NaCHO}_{2}\) to give this pH? (Assume that volumes are additive.)

Step-by-Step Solution

Answer

Answer: To prepare a buffer with a pH of 3.0, the concentration of the conjugate base (CHO₂⁻) must be about 0.1778 times the concentration of the weak acid (HCHO₂). For a 1-L solution of 0.139 M NaCHO₂: You will need 0.781 moles of HCHO₂. For 350 mL of 0.159 M HCHO₂: You will need 0.685 grams of NaCHO₂. For 1 L of 0.500 M NaCHO₂ with a 0.236 M HCHO₂ solution: You will need 11.90 L of the 0.236 M HCHO₂ solution.

Step 1: Write down the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by: \(p\mathrm{H} = p\mathrm{K}_{a} + \log \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\) where \(\mathrm{pH}\) is the pH of the buffer, \(p\mathrm{K}_{a}\) is the \(p\mathrm{K}_{a}\) value of the weak acid, \([\mathrm{A}^{-}]\) is the concentration of the conjugate base, and \([\mathrm{HA}]\) is the concentration of the weak acid.

Step 2: Calculate the pKa of the weak acid

The weak acid is \(\mathrm{HCHO}_{2}\) (formic acid), which has a \(p\mathrm{K}_{a}\) the value of 3.75.

Step 3: Find the ratio of acid to base concentrations

Using the Henderson-Hasselbalch equation and the given pH (3.0), we can find the ratio of acid to base concentrations: \(3.0 = 3.75 + \log \frac{[\mathrm{CHO}_{2}^{-}]}{[\mathrm{HCHO}_{2}]}\) Solve this equation for the ratio of acid to base concentrations: \(\log \frac{[\mathrm{CHO}_{2}^{-}]}{[\mathrm{HCHO}_{2}]} = 3.0 - 3.75\) \(\frac{[\mathrm{CHO}_{2}^{-}]}{[\mathrm{HCHO}_{2}]} = 10^{-0.75}\) The ratio is approximately \(0.1778\). This means that the concentration of the conjugate base ([CHO\(_2^{-}\)]) must be about 0.1778 times the concentration of the weak acid ([HCHO\(_2\)]) to achieve a pH of 3.0. Now, we can use this ratio to solve the remaining parts of the exercise:

Step 4: Moles of HCHO\(_2\) needed for a 1-L solution of 0.139 M NaCHO\(_2\)

We are given the concentration of NaCHO\(_2\), which corresponds to the concentration of the conjugate base, [CHO\(_2^{-}\)]. Using the ratio, we can find the concentration of the weak acid: \([\mathrm{HCHO}_{2}] = \frac{[\mathrm{CHO}_{2}^{-}]}{0.1778} = \frac{0.139}{0.1778} = 0.781 \mathrm{M}\) Now, multiply this concentration by the volume to obtain the number of moles of HCHO\(_2\) needed: Moles of \(\mathrm{HCHO}_{2} = 0.781 \mathrm{M} \times 1.00 \mathrm{L} = 0.781 \mathrm{mol}\)

Step 5: Grams of NaCHO\(_2\) needed for 350 mL of 0.159 M HCHO\(_2\)

First, find the moles of HCHO\(_2\) in 350 mL of a 0.159 M solution: Moles of \(\mathrm{HCHO}_{2} = 0.159 \mathrm{M} \times 0.350 \mathrm{L} = 0.0557 \mathrm{mol}\) Now, using the ratio, find the moles of NaCHO\(_2\) (which corresponds to moles of CHO\(_2^{-}\)): Moles of \(\mathrm{NaCHO}_{2} = 0.0557 \mathrm{mol} \times 0.1778 = 0.00992 \mathrm{mol}\) Finally, convert the moles of NaCHO\(_2\) to grams: Grams of \(\mathrm{NaCHO}_{2} = 0.00992 \mathrm{mol} \times (23.0 + 12.01 + 16.0 \mathrm{~g/mol}) = 0.685 \mathrm{g}\)

Step 6: Volume of 0.236 M HCHO\(_2\) needed for 1 L of 0.500 M NaCHO\(_2\)

First, find the moles of NaCHO\(_2\) in 1 L of 0.500 M solution: Moles of \(\mathrm{NaCHO}_{2} = 0.500 \mathrm{M} \times 1.00 \mathrm{L} = 0.500 \mathrm{mol}\) Now, using the ratio, find the moles of HCHO\(_2\) needed: Moles of \(\mathrm{HCHO}_{2} = 0.500 \mathrm{mol} \times \frac{1}{0.1778} = 2.81 \mathrm{mol}\) Finally, find the volume of the 0.236 M HCHO\(_2\) solution needed to have 2.81 moles of HCHO\(_2\): Volume of \(\mathrm{HCHO}_{2} \mathrm{~solution} = \frac{2.81 \mathrm{mol}}{0.236 \mathrm{M}} = 11.90 \mathrm{L}\)

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a crucial tool in chemistry for understanding how buffers work. It shows us how the pH of a solution is related to the concentration of an acid and its conjugate base. Using this equation, we can predict and adjust the pH of a buffer system effectively.

The equation is expressed as:
\[pH = pK_a + \log \frac{[A^-]}{[HA]}\]

In this equation:

• \(pH\) is the measure of the solution's acidity.
• \(pK_a\) is the acid dissociation constant, a special number that tells us how well an acid gives up its protons in solution.
• \([A^-]\) is the concentration of the conjugate base.
• \([HA]\) is the concentration of the weak acid.

This equation helps us to keep the pH of the buffer within a specific range by adjusting the concentrations of the acid and base.

Knowing the Henderson-Hasselbalch equation allows scientists to design buffer solutions that maintain desired pH levels in various chemical and biological applications.

pH calculation

Calculating the pH of a solution is a fundamental aspect of managing chemical reactions and biological processes. pH is a logarithmic scale used to specify the acidity or basicity of an aqueous solution. The pH scale typically ranges from 0 to 14.

Here's how you can think about it:

• If pH is below 7, the solution is acidic.
• If pH is 7, the solution is neutral.
• If pH is above 7, the solution is basic.

The pH of a solution can be calculated using the formula:
\[pH = -\log[H^+]\]
This formula tells us the pH depends on the concentration of hydrogen ions in the solution.

In the context of buffers, however, we often use the Henderson-Hasselbalch equation to calculate pH as it accounts for both the presence of a weak acid and its conjugate base, allowing us to predict how changes in concentration will affect the pH.

Acid-Base Equilibrium

Acid-base equilibrium is a central concept in chemistry related to how acids and bases interact in a solution to reach a state of balance. In a solution, the acid donates protons (H extsuperscript{+} ions), and the base accepts them, creating a balance called equilibrium.

Here are some key points about acid-base equilibrium:

• This balance is described using the equilibrium constant (K extsubscript{a}), specific to each acid. This constant indicates the extent of dissociation of the acid.
• The larger the K extsubscript{a}, the stronger the acid and the more it dissociates in the solution.
• Equilibrium is dynamic, meaning the forward and reverse reactions happen simultaneously at equal rates.

Understanding equilibrium helps us to predict the behavior of acids and bases in reactions, calculate pH, and design effective buffer systems. The Henderson-Hasselbalch equation we discussed earlier stems from these equilibrium principles and allows us to maintain the desired pH.

Formic Acid

Formic acid, also known by its chemical formula HCHO extsubscript{2}, is a simple carboxylic acid. It's quite common in chemistry and has some unique properties.

Some key characteristics of formic acid:

• It is the simplest carboxylic acid, making it an excellent example in teaching acid-base concepts.
• Has a strong, pungent smell and is naturally found in ant venom, giving it the name "formic."
• Has a relatively high \(pK_a = 3.75\), indicating the dissociation level when compared with other acids, which reflects its moderate acidity.

Formic acid is often used in buffer solutions thanks to its ability to maintain stable pH levels, making it incredibly useful in both industrial and laboratory settings.

When formic acid and its conjugate base, CHO extsubscript{2} extsuperscript{-}, are used in a buffer solution, they help resist changes in pH when small amounts of acids or bases are introduced, which is vital for precise scientific experiments.

Solution Concentration

Solution concentration is a measure of how much solute is dissolved in a given amount of solvent. It's crucial in the context of buffers, as the concentration of the acid and the base influences the buffer's capacity to maintain a stable pH.

There are several ways to express concentration, but one of the most common is molarity ( extbf{M}):

• Molarity is defined as moles of solute per liter of solution ( extbf{mol/L}).
• Concentration directly affects the strength of a buffer solution. Higher concentrations typically mean a stronger buffer capable of resisting pH changes.
• Understanding the concentration of the components in a buffer is essential when preparing solutions with specific pH requirements.

Being able to calculate and adjust concentrations accurately is key in not only creating effective buffers but also in many other fields such as pharmacology, geology, and environmental science.

In our exercises, concentrations help determine how much of each reactant we need to achieve the desired chemical balance and maintain the proper pH level.