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Chapter 16: Problem 78

In what volume ratio should you mix \(1.0 \mathrm{M}\) solutions of \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{NH}_{3}\) to produce a buffer solution having \(\mathrm{pH}=9.80\) ?

Short Answer

Expert verified
Mix 3.55 volumes of NH鈧� for every 1 volume of NH鈧凜l.

Step by step solution

01

Understand the Buffer Composition

A buffer solution consists of a weak base and its conjugate acid. In this exercise, \(\mathrm{NH}_3\) acts as the weak base and \(\mathrm{NH}_4Cl\) acts as the conjugate acid.
02

Apply the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is \( \text{pH} = \text{pKa} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \). For \(\mathrm{NH}_3\) and \ \mathrm{NH}_4^+ \, the \(\text{pKa}\) is 9.25.
03

Substitute Known Values

Given \( \text{pH} = 9.80 \) and \( \text{pKa} = 9.25 \), substitute them into the Henderson-Hasselbalch equation: \(9.80 = 9.25 + \log \left( \frac{[\mathrm{NH}_3]}{[\mathrm{NH}_4Cl]} \right)\).
04

Solve for Base to Acid Ratio

Isolate the log term: \( \log \left( \frac{[\mathrm{NH}_3]}{[\mathrm{NH}_4Cl]} \right) = 9.80 - 9.25 = 0.55 \). Then find the antilog by taking \(10^{0.55}\) to obtain the base to acid ratio: \(\frac{[\mathrm{NH}_3]}{[\mathrm{NH}_4Cl]} = 10^{0.55} \). Calculating gives approximately 3.55.
05

Interpret the Ratio

The ratio \( \frac{[\mathrm{NH}_3]}{[\mathrm{NH}_4Cl]} \approx 3.55 \) means that you need 3.55 volumes of \ \mathrm{NH}_3 \ for every 1 volume of \ \mathrm{NH}_4Cl \ to achieve the desired pH.

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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 a fundamental formula used to calculate the pH of a buffer solution. It has the form:
  • \( \text{pH} = \text{pKa} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \)
This equation helps us understand how the concentrations of a weak acid or base and its conjugate affect the pH of the solution.
This relationship is essential for designing buffer solutions, which are used to resist changes in pH when adding small amounts of acid or base.
When using this equation, determine the respective concentrations of the weak base and the conjugate acid present in the buffer solution.
Weak Base
A weak base is a substance that partially accepts protons in solution. Unlike strong bases, they do not dissociate completely.
In the context of buffers, a weak base like \( \text{NH}_3 \) can react with added acids to help maintain the pH.
  • \( \text{NH}_3 \) can form \( \text{NH}_4^+ \) by accepting a proton from the solution.
  • This reaction between \( \text{NH}_3 \) and acids is reversible and helps stabilize pH.
Using a weak base is crucial because the reversible reactions allow for efficient regulation of pH changes, providing a stable environment for various applications, such as in biological and chemical systems.
Conjugate Acid
The term conjugate acid refers to the species formed when a base accepts a proton.
Using the example of ammonia \( \text{NH}_3 \), the conjugate acid is ammonium \( \text{NH}_4^+ \).
This relationship is vital in buffer systems; both the weak base and its conjugate acid co-exist to resist pH changes.
  • As \( \text{NH}_3 \) gains a proton, it forms \( \text{NH}_4^+ \).
  • Conversely, \( \text{NH}_4^+ \) can lose a proton to reform \( \text{NH}_3 \).
This duality allows the solution to absorb excess \( \text{H}^+ \) or \( \text{OH}^- \) ions, minimizing the pH fluctuation, essential for maintaining ideal conditions in many scientific and industrial processes.
pH Calculation
The pH calculation in a buffer system involves determining the hydrogen ion concentration, which is influenced by the base and acid concentrations.
In buffers, pH reflects the balance between weak base and its conjugate acid. The Henderson-Hasselbalch equation comes into play here.
  • With \( \text{pH} = 9.80 \), the calculation started with known \( \text{pKa} \).
  • Adjust base:acid ratio using the formula to maintain this pH.
By plugging values into the formula, solve for the necessary ratio of \([\text{Base}]:[\text{Acid}]\). For instance, to achieve \( \text{pH} = 9.80 \), a ratio like \( 3.55:1 \) is determined for \( \text{NH}_3 \) and \( \text{NH}_4\text{Cl} \). This ratio is crucial for the buffer system's effectiveness.

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Most popular questions from this chapter

Which of the following compounds are more soluble in acidic solution than in pure water? Write a balanced net ionic equation for each dissolution reaction. (a) \(\mathrm{AgBr}\) (b) \(\mathrm{CaCO}_{3}\) (c) \(\mathrm{Ni}(\mathrm{OH})_{2}\) (d) \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

Calculate the \(\mathrm{pH}\) in a solution prepared by dissolving \(0.10 \mathrm{~mol}\) of solid \(\mathrm{NH}_{4} \mathrm{Cl}\) in \(0.500 \mathrm{~L}\) of \(0.40 \mathrm{M} \mathrm{NH}_{3}\). Assume that there is \(\mathrm{no}\) volume change.

Prior to having an X-ray exam of the upper gastrointestinal tract, a patient drinks an aqueous suspension of solid \(\mathrm{BaSO}_{4}\). (Scattering of \(\mathrm{X}\) rays by barium greatly enhances the quality of the photograph.) Although \(\mathrm{Ba}^{2+}\) is toxic, ingestion of \(\mathrm{BaSO}_{4}\) is safe because it is quite insoluble. If a saturated solution prepared by dissolving solid \(\mathrm{BaSO}_{4}\) in water has \(\left[\mathrm{Ba}^{2+}\right]=1.05 \times 10^{-5} \mathrm{M}\), what is the value of \(K_{\mathrm{sp}}\) for \(\mathrm{BaSO}_{4} ?\)

APPLY 16.4 Calculate the \(\mathrm{pH}\) of a solution prepared by mixing equal volumes of \(0.20 \mathrm{M}\) methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}, K_{\mathrm{b}}=3.7 \times 10^{-4}\right)\) and \(0.60 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\).

A \(50.0 \mathrm{~mL}\) sample of \(0.250 \mathrm{M}\) ammonia \(\left(\mathrm{NH}_{3}, K_{\mathrm{b}}=1.8 \times 10^{-5}\right)\) is titrated with \(0.250 \mathrm{M} \mathrm{HNO}_{3} .\) Calculate the \(\mathrm{pH}\) after the addition of each of the following volumes of acid: (a) \(0.0 \mathrm{~mL}\) (b) \(25.0 \mathrm{~mL}\) (c) \(50.0 \mathrm{~mL}\) (d) \(60.0 \mathrm{~mL}\)
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