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Chapter 8: Problem 249

\(\mathrm{pOH}\) of an aqueous solution is \(5.0 .\) If the conc. of \(\left[\mathrm{H}^{+}\right]\) in that solution is \(10^{-x} \mathrm{M}\), the value of \(\mathrm{x}\) is

Short Answer

Expert verified
The value of \( x \) is 9.0.

Step by step solution

01

Understanding the pOH Definition

The pOH of a solution is defined as the negative logarithm (base 10) of the hydroxide ion concentration: \( \text{pOH} = -\log [\text{OH}^-] \). In this problem, we are given \( \text{pOH} = 5.0 \).
02

Calculate [OH鈦籡

To find the hydroxide ion concentration \([\text{OH}^-]\), use the definition of pOH: \[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-5.0} \]. This means \([\text{OH}^-] = 10^{-5} \text{ M}\).
03

Relationship Between pH and pOH

For aqueous solutions at 25掳C, the sum of pH and pOH is always 14: \( \text{pH} + \text{pOH} = 14 \). Given \( \text{pOH} = 5.0 \), calculate \( \text{pH} \) as follows: \( \text{pH} = 14 - 5.0 = 9.0 \).
04

Calculate [H鈦篯 from pH

The hydrogen ion concentration \([\text{H}^+]\) can be calculated from the pH: \[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-9.0} \].
05

Relate [H鈦篯 to Given Expression

We are given that \([\text{H}^+] = 10^{-x} \text{ M}\). Since from our calculation \([\text{H}^+] = 10^{-9.0} \text{ M}\), this implies that \( x = 9.0 \).

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Key Concepts

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

Understanding Hydroxide Ion Concentration
The hydroxide ion concentration is a crucial part of understanding the behavior of alkaline or basic solutions. Hydroxide ions, denoted as \([ ext{OH}^-]\), contribute to a solution's basicity.
In terms of the pOH scale, which measures how basic a solution is, it is defined as the negative log of this concentration. In mathematical form, pOH is expressed as:
  • \( \text{pOH} = -\log [\text{OH}^-] \)
Given the pOH of 5.0 in our exercise, we find the hydroxide ion concentration by reversing the logarithmic process. We do this by calculating:
  • \([\text{OH}^-] = 10^{\text{-pOH}} = 10^{-5.0} = 10^{-5} \text{ M}\)
This helps determine how basic the aqueous solution actually is.
Exploring the Relationship Between pH and pOH
The relationship between pH and pOH is foundational for understanding aqueous solutions, especially at a temperature of 25掳C. The sum of pH and pOH for such solutions is always 14. This relationship is described by the equation:
  • \( \text{pH} + \text{pOH} = 14 \)
This makes it easy to convert between pOH and pH. Given our previous value of \( \text{pOH} = 5.0 \), we can calculate pH by simply subtracting from 14, giving:
  • \( \text{pH} = 14 - 5.0 = 9.0 \)
Understanding this equation allows one to switch between measuring acidity and basicity.
Determining Hydrogen Ion Concentration
Hydrogen ion concentration is a core concept when discussing the acidity of a solution. It is denoted as \([\text{H}^+]\). The pH, which tells us how acidic a solution is, is calculated with the formula:
  • \( \text{pH} = -\log [\text{H}^+] \)
Therefore, to find \([\text{H}^+]\) given the pH, we reverse the logarithm:
  • \([\text{H}^+] = 10^{-\text{pH}} \)
From our determined pH of 9.0, the calculation becomes:
  • \([\text{H}^+] = 10^{-9.0} \text{ M}\)
This value is critical for understanding the solution's acidic nature and, in this case, confirms the calculation, concluding \( x = 9.0 \) if \([\text{H}^+] = 10^{-x} \text{ M}\).

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Most popular questions from this chapter

The solubility product of \(\mathrm{PbI}_{2}\) is \(7.47 \times 10^{-9}\) at \(15^{\circ} \mathrm{C}\) and \(1.39 \times 10^{8}\) at \(25^{\circ} \mathrm{C}\). The molar heat of solution of \(\mathrm{PbI}_{2}\) is (use \(\log 1.86=0.2695\) ) (a) \(44.29 \mathrm{~kJ} / \mathrm{mol}\) (b) \(46.25 \mathrm{~kJ} / \mathrm{mol}\) (c) \(29.37 \mathrm{~kJ} / \mathrm{mol}\) (d) \(21.15 \mathrm{~kJ} / \mathrm{mol}\)

Which of the following statements is/are correct about the ionic product of water ? (a) At \(25^{\circ} \mathrm{C}, \mathrm{K}\) (dissociation constant of water \()>\mathrm{K}_{\mathrm{w}}\) (ionic product of water) (b) \(K_{w}\) of boiling water is greater than \(10^{-14}\). (c) Ionic product of water at \(25^{\circ} \mathrm{C}\) is \(10^{-14}\) (d) \(\mathrm{pH}+\mathrm{pOH}=\mathrm{pK}_{\mathrm{w}^{\circ}}\)

If molar concentrations of two weak acids are the same, their relative strengths can be compared by (a) \(\frac{\alpha_{1}}{\alpha_{2}}\) (b) \(\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}\) (c) \(\sqrt{\mathrm{K}_{1} / \mathrm{K}_{2}}\) (d) \(\frac{\left[\mathrm{H}^{+}\right]_{1}}{\left[\mathrm{H}^{+}\right]_{2}}\)

A buffer solution can be prepared from a mixture of (a) \(\mathrm{CH}_{3} \mathrm{COONa}\) and \(\mathrm{CH}_{3} \mathrm{COOH}\) in water (b) \(\mathrm{CH}_{3} \mathrm{COONa}\) and \(\mathrm{HCl}\) in water under certain conditions (c) \(\mathrm{NH}_{4} \mathrm{OH}\) and \(\mathrm{NH}_{4} \mathrm{Cl}\) in water (d) \(\mathrm{NaCl}\) and \(\mathrm{HCl}\) in water

Which of the following are wrong statements? (a) \(\mathrm{K}_{\mathrm{i}}\) is always constant and equal to \(10^{-14}\) (b) \(\mathrm{pH}+\mathrm{pOH}=\mathrm{pK}_{w}\) at all temperature (c) Salts of strong acid and strong base undergo hydrolysis (d) Addition of \(\mathrm{CH}_{3} \mathrm{COONa}\) to acetic acid solution decreases the \(\mathrm{pH}\) of solution of acetic acid.
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