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Chapter 16: Problem 62

The \(\mathrm{pH}\) of a solution of \(\mathrm{HN}_{3}\left(K_{\mathrm{a}}=1.9 \times 10^{-5}\right)\) and \(\mathrm{NaN}_{3}\) is \(4.86\). What is the molarity of \(\mathrm{NaN}_{3}\) if the molarity of \(\mathrm{HN}_{3}\) is \(0.016 \mathrm{M}\) ?

Short Answer

Expert verified
The molarity of \(\mathrm{NaN}_3\) is \(0.022 \, \mathrm{M}\).

Step by step solution

01

Understanding the Problem

We have a solution containing the weak acid \(\mathrm{HN}_3\) and its conjugate base \(\mathrm{NaN}_3\). We are given the \(\mathrm{pH}\) of the solution, which is \(4.86\), and the molarity of \(\mathrm{HN}_3\), which is \(0.016\,\mathrm{M}\). Our goal is to find the molarity of \(\mathrm{NaN}_3\).
02

Determine the \\([\mathrm{H}^+]\\) Concentration

First, we convert \(\mathrm{pH}\) to \([\mathrm{H}^+]\) concentration using the formula \([\mathrm{H}^+] = 10^{-\mathrm{pH}}\). For \(\mathrm{pH} = 4.86\), we find \([\mathrm{H}^+] = 10^{-4.86} \approx 1.38 \times 10^{-5} \, \mathrm{M}\).
03

Using the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation relates \(\mathrm{pH}\), the acid dissociation constant \(K_a\), and the concentrations of the acid and its conjugate base:\[ \mathrm{pH} = \mathrm{p}K_a + \log{\left( \frac{[\mathrm{A}^-]}{[\mathrm{HA}]} \right)} \]\[ \mathrm{p}K_a = -\log{K_a} = -\log{(1.9 \times 10^{-5})} \approx 4.72 \]
04

Calculate the Ratio \\([\mathrm{A}^-]/[\mathrm{HA}]\\)

Rearrange the Henderson-Hasselbalch equation to find the ratio:\[ 4.86 = 4.72 + \log{\left( \frac{[\mathrm{NaN}_3]}{0.016} \right)} \]\[ 0.14 = \log{\left( \frac{[\mathrm{NaN}_3]}{0.016} \right)} \]Take the antilog:\[ 10^{0.14} = \frac{[\mathrm{NaN}_3]}{0.016} \]
05

Solve for \\([\mathrm{NaN}_3]\\)

Calculate the right-hand side:\[ 10^{0.14} \approx 1.38 \]So, \[ 1.38 = \frac{[\mathrm{NaN}_3]}{0.016} \]Multiply both sides by \(0.016\):\[ [\mathrm{NaN}_3] \approx 0.016 \times 1.38 \approx 0.02208 \, \mathrm{M} \]
06

Round the Final Answer

Round the molarity of \(\mathrm{NaN}_3\) to an appropriate number of significant figures based on the given data, which is \(0.022 \, \mathrm{M} \).

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Key Concepts

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

Acid-Base Chemistry
Acid-base chemistry explores the behavior of acids and bases in solutions. It involves understanding how these substances interact, react, and change in different conditions. Acids, like \( H_3\), donate protons (\([ H^+]\)), while bases, like \( NaN_3\), accept them. Breaking this down further:
  • Acids add hydrogen ions to solutions, making them more acidic.
  • Bases remove hydrogen ions, increasing the solution's pH, making it more basic.
In this exercise, we observe an equilibrium between the weak acid \( HN_3\) and its conjugate base \( NaN_3\), forming a buffer system that resists changes in \( pH\). Understanding this dynamic relationship helps in studying how to adjust or stabilize \( pH\) in various chemical processes.
pH Calculation
Calculating \( pH\) is a fundamental aspect of understanding acid-base equilibria. The \( pH\) scale measures how acidic or basic a solution is, ranging typically from 0 to 14. To find \( pH\), you use the formula:
  • \( pH = -\log{[H^+]}\)
Where \([ H^+]\) is the concentration of hydrogen ions in moles per liter (M). In this particular problem, we know the \( pH\) of the solution is 4.86, which can be converted back to find \([ H^+]\). This helps us understand the acidity level of the solution. Calculating \( pH\) provides key insights into the balance of acid and base in any mixture, which is essential in many scientific and industrial processes.
Weak Acid and Conjugate Base
The relationship between a weak acid and its conjugate base forms the backbone of buffer systems. A weak acid doesn't completely ionize in water, meaning only a portion of it turns into hydrogen ions. For \( HN_3\), this acid partially disassociates in solution.
  • **Weak Acid:** Only partially disassociates.
  • **Conjugate Base:** What the acid turns into after losing a proton.
In this exercise, \( NaN_3\) acts as the conjugate base, resulting from \( HN_3\)'s partial ionization. The presence of both the weak acid and the conjugate base in the solution helps maintain the \( pH\) through an equilibrium process, which resists drastic changes if small amounts of acids or bases are added.
Acid Dissociation Constant
The acid dissociation constant, represented as \( K_a\), measures the strength of an acid in solution. It indicates how well an acid disassociates into its ions. A high \( K_a\) means a strong acid, while a low \( K_a\) signifies a weak acid.
  • The formula for \( K_a\) is: \([H^+][A^-]/[HA]\)
In this problem, the \( K_a\) for \( HN_3\) is given as \( 1.9 \times 10^{-5}\), indicating it is a weak acid. This value is crucial for calculating \( pK_a\), another term used in the Henderson-Hasselbalch equation. \( pK_a = -\log{K_a}\), and in this case, it's approximately 4.72. This conversion is essential for determining the \( pH\) of the buffer solution, linking the strength of the acid to the pH scale.

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Most popular questions from this chapter

Use Le Ch芒telier's principle to predict whether the solubility of \(\mathrm{BaF}_{2}\) will increase, decrease, or remain the same on addition of each of the following substances: (a) \(\mathrm{HCl}\) (b) \(\mathrm{KF}\) (c) \(\mathrm{NaNO}_{3}\) (d) \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\)

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 \(\mathrm{BaSO}_{4} ?\)

Calculate the molar solubility of \(\operatorname{SrF}_{2}\) in: (a) \(0.010 \mathrm{M} \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\) (b) \(0.010 \mathrm{M} \mathrm{NaF}\)

Which of the following gives a buffer solution when equal volumes of the two solutions are mixed? (a) \(0.10 \mathrm{M} \mathrm{HF}\) and \(0.10 \mathrm{M} \mathrm{NaF}\) (b) \(0.10 \mathrm{M} \mathrm{HF}\) and \(0.10 \mathrm{M} \mathrm{NaOH}\) (c) \(0.20 \mathrm{M} \mathrm{HF}\) and \(0.10 \mathrm{M} \mathrm{NaOH}\) (d) \(0.10 \mathrm{M} \mathrm{HCl}\) and \(0.20 \mathrm{M} \mathrm{NaF}\)

What compound, if any, will precipitate when \(80 \mathrm{~mL}\) of \(1.0 \times 10^{-5} \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) is added to \(20 \mathrm{~mL}\) of \(1.0 \times 10^{-5} \mathrm{M}\) \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3} ?\)
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