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Chapter 14: Problem 50

Calculate the \(\left[\mathrm{H}^{+}\right]\) of each of the following solutions at \(25^{\circ} \mathrm{C}\) . Identify each solution as neutral, acidic, or basic. a. \(\left[\mathrm{OH}^{-}\right]=1.5 M\) b. \(\left[\mathrm{OH}^{-}\right]=3.6 \times 10^{-15} M\) c. \(\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-7} M\) d. \(\left[\mathrm{OH}^{-}\right]=7.3 \times 10^{-4} M\) Also calculate the pH and pOH of each of these solutions.

Short Answer

Expert verified
The calculated concentrations of \(\left[\mathrm{H}^{+}\right]\) and pH and pOH values are as follows: a. \(\left[\mathrm{H}^{+}\right] = 6.67 \times 10^{-15} M\), pH=14.18, pOH=-0.18, basic solution. b. \(\left[\mathrm{H}^{+}\right] = 2.77 \times 10^{-14} M\), pH=13.56, pOH=14.44, basic solution. c. \(\left[\mathrm{H}^{+}\right] = 1.0 \times 10^{-7} M\), pH=7, pOH=7, neutral solution. d. \(\left[\mathrm{H}^{+}\right] = 1.37 \times 10^{-12} M\), pH=11.86, pOH=3.14, basic solution.

Step by step solution

01

Calculate the \(\left[\mathrm{H}^{+}\right]\)

Using the \(K_w\) equation, we can calculate the \(\left[\mathrm{H}^{+}\right]\): \(\left[\mathrm{H}^{+}\right] = \dfrac{K_w}{\left[\mathrm{OH}^{-}\right]} = \dfrac{1 \times 10^{-14}}{1.5} = 6.67 \times 10^{-15} M\)
02

Determine the pH and pOH

Now, we will calculate the pH and pOH values: \(pH = -\log{(6.67 \times 10^{-15})}= 14.18\) \(pOH = -\log{(1.5)}= -0.18\) Since the pH value is greater than 7, it is a basic solution. b. \(\left[\mathrm{OH}^{-}\right]=3.6 \times 10^{-15} M\)
03

Calculate the \(\left[\mathrm{H}^{+}\right]\)

Similar to part a, we will calculate the \(\left[\mathrm{H}^{+}\right]\): \(\left[\mathrm{H}^{+}\right] = \dfrac{K_w}{\left[\mathrm{OH}^{-}\right]} = \dfrac{1 \times 10^{-14}}{3.60 \times 10^{-15}} = 2.77 \times 10^{-14} M\)
04

Determine the pH and pOH

Now, we will calculate the pH and pOH values: \(pH = -\log{(2.77 \times 10^{-14})}= 13.56\) \(pOH = -\log{(3.6 \times 10^{-15})}= 14.44\) Since the pH value is greater than 7, it is a basic solution. c. \(\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-7} M\)
05

Calculate the \(\left[\mathrm{H}^{+}\right]\)

Similar to part a, we will calculate the \(\left[\mathrm{H}^{+}\right]\): \(\left[\mathrm{H}^{+}\right] = \dfrac{K_w}{\left[\mathrm{OH}^{-}\right]} = \dfrac{1 \times 10^{-14}}{1 \times 10^{-7}} = 1.0 \times 10^{-7} M\)
06

Determine the pH and pOH

Now, we will calculate the pH and pOH values: \(pH = -\log{(1.0 \times 10^{-7})}= 7\) \(pOH = -\log{(1.0 \times 10^{-7})}= 7\) Since the pH value is equal to 7, it is a neutral solution. d. \(\left[\mathrm{OH}^{-}\right]=7.3 \times 10^{-4} M\)
07

Calculate the \(\left[\mathrm{H}^{+}\right]\)

Similar to part a, we will calculate the \(\left[\mathrm{H}^{+}\right]\): \(\left[\mathrm{H}^{+}\right] = \dfrac{K_w}{\left[\mathrm{OH}^{-}\right]} = \dfrac{1 \times 10^{-14}}{7.3 \times 10^{-4}} = 1.37 \times 10^{-12} M\)
08

Determine the pH and pOH

Now, we will calculate the pH and pOH values: \(pH = -\log{(1.37 \times 10^{-12})}= 11.86\) \(pOH = -\log{(7.3 \times 10^{-4})}= 3.14\) Since the pH value is greater than 7, it is a basic solution.

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Key Concepts

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

Acidic and Basic Solutions
Acidic and basic solutions are key concepts in understanding chemical reactions. Whether a solution is acidic, neutral, or basic depends on the concentration of hydrogen ions (H鈦�) in the solution.
A solution is considered
  • **Acidic** if its pH is less than 7. This indicates a higher concentration of hydrogen ions than hydroxide ions (OH鈦�).
  • **Neutral** if its pH is exactly 7. Here, the concentrations of H鈦� and OH鈦� are equal.
  • **Basic** if its pH is greater than 7. This shows a higher concentration of hydroxide ions compared to hydrogen ions.
Understanding these distinctions allows you to predict the behavior of different substances in a chemical environment. Calculating the pH helps in identifying whether a solution is acidic, basic, or neutral.
Hydrogen Ion Concentration
Hydrogen ion concentration (\[\text{[H}^+\text{]}\]) is critical to finding the acidity of the solution. It is directly related to the pH value through a simple relationship: **pH is the negative logarithm of the hydrogen ion concentration (pH = -log[H鈦篯).**
This equation helps us calculate how acidic a solution is:
  • Higher [H鈦篯 means a lower pH, indicating more acidity.
  • Lower [H鈦篯 means a higher pH, indicating less acidity and potentially a basic solution.
Use this knowledge to calculate pH from known hydrogen ion concentrations, which is essential for determining the nature of solutions.
Relationship Between pH and pOH
The relationship between pH and pOH is vital for understanding the balance between acidity and basicity. Both can be calculated using the ion product of water (\(K_w\)= 1 脳 10鈦宦光伌 at 25掳C).
This relationship is expressed as:
  • \(\text{pH} + \text{pOH} = 14\)
Knowing one allows you to easily find the other. For example, if you know the pH, subtracting it from 14 gives you the pOH.
This interconnection helps in balancing equations and understanding the nature of solutions in different chemical environments.

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Most popular questions from this chapter

Write the reaction and the corresponding \(K_{\mathrm{b}}\) equilibrium expression for each of the following substances acting as bases in water. a. aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) b. dimethylamine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\)

Calculate the concentration of all species present and the pH of a \(0.020-M\) HF solution.

Place the species in each of the following groups in order of increasing acid strength. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{S}, \mathrm{H}_{2} \mathrm{Se}\) (bond energies: \(\mathrm{H}-\mathrm{O}, 467 \mathrm{kJ} / \mathrm{mol}\) \(\mathrm{H}-\mathrm{S}, 363 \mathrm{kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{Se}, 276 \mathrm{kJ} / \mathrm{mol} )\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{FCH}_{2} \mathrm{CO}_{2} \mathrm{H}, \mathrm{F}_{2} \mathrm{CHCO}_{2} \mathrm{H}, \mathrm{F}_{3} \mathrm{CCO}_{2} \mathrm{H}\) c. \(\mathrm{NH}_{4}^{+}, \mathrm{HONH}_{3}^{+}\) d. \(\mathrm{NH}_{4}^{+}, \mathrm{PH}_{4}^{+}\) (bond energies: \(\mathrm{N}-\mathrm{H}, 391 \mathrm{kJ} / \mathrm{mol} ; \mathrm{P}-\mathrm{H},\) 322 \(\mathrm{kJ} / \mathrm{mol} )\) Give reasons for the orders you chose.

For propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right),\) determine the concentration of all species present, the \(\mathrm{pH},\) and the percent dissociation of a \(0.100-M\) solution.

Consider 1000 . mL of a \(1.00 \times 10^{-4}-M\) solution of a certain acid HA that has a \(K_{\text { a value equal to } 1.00 \times 10^{-4} . \text { How much }}\) water was added or removed (by evaporation) so that a solution remains in which 25.0\(\%\) of \(\mathrm{HA}\) is dissociated at equilibrium? Assume that HA is nonvolatile.
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