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Chapter 9: Problem 27

What mass of sodium formate must be added to \(500.0 \mathrm{~mL}\) of \(1.00 \mathrm{M}\) formic acid to produce a buffer solution that has a \(\mathrm{pH}\) of \(3.50\) ?

Short Answer

Expert verified
Add 19.04 g of sodium formate to the solution.

Step by step solution

01

Understand the Buffer System

We are dealing with a buffer solution made up of a weak acid and its conjugate base. In this case, formic acid ( HCOOH) and sodium formate ( HCOONa) form the buffer pair, where sodium formate acts as the conjugate base to formic acid.
02

Use the Henderson-Hasselbalch Equation

The pH of a buffer can be calculated using the Henderson-Hasselbalch equation:\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]We know the pH is 3.5, the concentration of formic acid \([\text{HA}]=1.00\,\mathrm{M}\), and need to find the concentration of sodium formate \([\text{A}^-]\). The \(\text{pKa}\) of formic acid is approximately 3.75.
03

Calculate the Concentration of Sodium Formate

Rearrange the Henderson-Hasselbalch equation to solve for \([\text{A}^-]\):\[ 3.50 = 3.75 + \log \left( \frac{[\text{A}^-]}{1.00} \right) \]Subtract 3.75 from both sides:\[ 3.50 - 3.75 = \log \left( \frac{[\text{A}^-]}{1.00} \right) \]This simplifies to:\[ -0.25 = \log \left( [\text{A}^-] \right) \]To remove the logarithm, take the antilog (10 raised to the power of both sides):\[ 10^{-0.25} = [\text{A}^-] \]This gives \([\text{A}^-] \approx 0.56\,\mathrm{M}\).
04

Calculate the Mass of Sodium Formate Needed

The concentration \([\text{A}^-]\) is 0.56 M, meaning we need 0.56 moles per liter. For 500 mL (0.5 L), the moles of sodium formate required are:\[0.56 \times 0.5 = 0.28 \text{ moles} \]The molar mass of sodium formate (\(\mathrm{HCOONa}\)) is approximately 68.01 g/mol. Therefore, the mass needed is:\[0.28 \times 68.01 \approx 19.04\, \mathrm{g}\]
05

Conclusion

To prepare the buffer solution with a pH of 3.50, add approximately 19.04 grams of sodium formate to 500 mL of 1.00 M formic acid.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is crucial for understanding buffer solutions. It offers a straightforward way to determine the pH of a solution containing a weak acid and its conjugate base. The equation is expressed as: \[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]Where:
  • pH is the measure of acidity of the buffer.
  • pKa is the acid dissociation constant, a fixed value specific to each weak acid.
  • [A^-] represents the concentration of the conjugate base.
  • [HA] is the concentration of the weak acid.
This equation takes into account the balance between the acid and base in a buffer, offering insight into how the components affect the pH. By rearranging the equation, you can solve for any variable, given the other known quantities. It exemplifies the logarithmic connection between hydrogen ion concentration and pH, making it invaluable for buffer preparation and analysis.
weak acid and conjugate base
A buffer solution typically consists of a weak acid and its conjugate base. In our exercise, formic acid (\( \text{HCOOH} \)) serves as the weak acid, while sodium formate (\( \text{HCOONa} \)) acts as its conjugate base. Why weak acids?
  • Weak acids partially dissociate in solution, which helps maintain a balance between the undissociated acid (\( \text{HA} \)) and the dissociated ions (\( \text{H}^+ \) and \( \text{A}^- \)).
  • This partial dissociation is what gives buffers their ability to resist significant changes in pH when either acid or base is added.
The role of conjugate bases:
  • The conjugate base is what remains after a weak acid donates its hydrogen ion.
  • It can recombine with \( \text{H}^+ \) ions, minimizing pH changes by controlling the ratio of \( [\text{A}^-] \) to \( [\text{HA}] \).
This relationship between the weak acid and its conjugate base is essential for the stability of buffers, allowing them to function effectively in chemical reactions and biological systems.
pH calculation
Calculating the pH of a buffer solution is a fundamental step in understanding its chemistry. For buffers built from a weak acid and its conjugate base, the pH is not determined directly from the concentration of \( \text{H}^+ \) ions. Instead, it involves using the Henderson-Hasselbalch equation. Steps to calculate pH:
1. **Identify and know the values needed:**
- The **pKa** of the weak acid. - Concentrations of the weak acid \( ([\text{HA}]) \) and the conjugate base \( ([\text{A}^-]) \).2. **Plug the known values into the Henderson-Hasselbalch equation**:
\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]3. **Solve for pH:**
Substitute the available values and calculate the pH.This equation simplifies the calculation by translating concentration ratios and dissociation constants into a logarithmic scale, leading to an efficient way of determining how acidic or basic a solution is likely to be. By following the calculated steps, you can predict and manipulate the behavior of a buffer system successfully.

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Most popular questions from this chapter

Briefly explain why there is no term in an equilibriumconstant expression for water or for a pure solid, even though one (or both) appears in the balanced net ionic equation for the equilibrium.

Define buffer capacity.

Consider solutions prepared by (a) dissolving \(8.00 \mathrm{mmol}\) of \(\mathrm{NaOAc}\) in \(200 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HOAc}\). (b) adding \(100 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{NaOH}\) to \(100 \mathrm{~mL}\) of \(0.175 \mathrm{M} \mathrm{HOAc}\). (c) adding \(40.0 \mathrm{~mL}\) of \(0.1200 \mathrm{M} \mathrm{HCl}\) to \(160.0 \mathrm{~mL}\) of \(0.0420 \mathrm{M} \mathrm{NaOAc}\). In what respects do these solutions resemble one another? How do they differ?

Is the following statement true or false or both? Define your answer with equations, examples, or graphs. "A buffer maintains the \(\mathrm{pH}\) of a solution constant."

Which has the greater buffer capacity: (a) a mixture containing \(0.100 \mathrm{~mol}\) of \(\mathrm{NH}_{3}\) and \(0.200 \mathrm{~mol}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) or (b) a mixture containing \(0.0500 \mathrm{~mol}\) of \(\mathrm{NH}_{3}\) and \(0.100 \mathrm{~mol}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) ?
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