Problem 9
Question
Calculate \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for each solution:(a) \(0.00165 \mathrm{M} \mathrm{HNO}_{3} ;\) (b) \(0.0087 \mathrm{M} \mathrm{KOH} ;\) (c) \(0.00213 \mathrm{M}\) \(\operatorname{Sr}(\mathrm{OH})_{2} ;(\mathrm{d}) 5.8 \times 10^{-4} \mathrm{M} \mathrm{HI}\)
Step-by-Step Solution
Verified Answer
\(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) concentrations for each soultion are: (a) \(0.00165 M\) and \(6.06 \times 10^{-12} M\); (b) \(6.06 \times 10^{-12} M\) and \(0.0087 M\); (c) \(2.35 \times 10^{-12} M\) and \(0.00426 M\); (d) \(5.8 \times 10^{-4} M\) and \(1.72 \times 10^{-11} M\).
1Step 1: Solve for (a): 0.00165 M HNO3
This is a strong acid. For every mole of HNO3, one mole of \(\mathrm{H}_{3} \mathrm{O}^{+}\) is produced. So, the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions is \(0.00165 M\). Knowing that \(KW=[\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{OH}^{-}]=1.0 \times 10^{-14} M^{2} \), we can find the concentration of \(\mathrm{OH}^{-}\) ions to be \( \frac{1.0 \times 10^{-14} M^2}{0.00165M} = 6.06 \times 10^{-12} M\).
2Step 2: Solve for (b): 0.0087 M KOH
KOH is a strong base. One mole of KOH results in one mole of \(\mathrm{OH}^{-}\) ions. So, the concentration of \(\mathrm{OH}^{-}\) is \(0.0087M\). Using the water ion product, we get the \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration to be \( \frac{1.0 \times 10^{-14} M^2}{0.0087M} = 1.15 \times 10^{-12} M \).
3Step 3: Solve for (c): 0.00213 M Sr(OH)2
This is a strong base with two hydroxyl ions for each formula unit. This means we have \(0.00213 M \times 2 = 0.00426 M\) of \(\mathrm{OH}^{-} \). The \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration is thus \( \frac{1.0 \times 10^{-14} M^2}{0.00426M} = 2.35 \times 10^{-12} M \).
4Step 4: Solve for (d): 5.8 x 10^{-4} M HI
HI is a strong acid. For each mole of HI, one mole of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions is produced. This makes the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) to be \( 5.8 \times 10^{-4} M \). Now, compute for the \(\mathrm{OH}^{-}\) concentration giving \( \frac{1.0 \times 10^{-14} M^2}{5.8 \times 10^{-4}M} = 1.72 \times 10^{-11} M \).
Key Concepts
Hydronium Ion ConcentrationHydroxide Ion ConcentrationStrong Acids and Bases
Hydronium Ion Concentration
Hydronium ion concentration reflects the acidic nature of a solution. In aqueous solutions, an acid donates protons (H⁺), which then bond with water molecules (H₂O) to form hydronium ions (H₃O⁺).
This process increases the hydronium concentration in the solution, indicating stronger acidity.
For strong acids, this relationship is straightforward. Each molecule of a strong acid like nitric acid (HNO₃) or hydroiodic acid (HI) dissociates completely in water. This means that the concentration of the acid equals the concentration of hydronium ions in the solution.
Let's consider nitric acid (HNO₃) at a concentration of 0.00165 M:
This process increases the hydronium concentration in the solution, indicating stronger acidity.
For strong acids, this relationship is straightforward. Each molecule of a strong acid like nitric acid (HNO₃) or hydroiodic acid (HI) dissociates completely in water. This means that the concentration of the acid equals the concentration of hydronium ions in the solution.
Let's consider nitric acid (HNO₃) at a concentration of 0.00165 M:
- Because it is a strong acid, it fully dissociates, creating 0.00165 M hydronium ions.
Hydroxide Ion Concentration
The hydroxide ion concentration in a solution signifies the basic nature of the solution. When a base dissolves in water, it usually generates hydroxide ions (OH⁻).
Strong bases like potassium hydroxide (KOH) or strontium hydroxide \((\text{Sr(OH)}_2\)) release hydroxide ions in a similar way strong acids release hydronium ions. This means each unit of these bases will dissociate completely to release hydroxide ions.
For KOH with a concentration of 0.0087 M:
The presence of these ions dictates the basicity of the solutions and influences acid-base equilibria.
Strong bases like potassium hydroxide (KOH) or strontium hydroxide \((\text{Sr(OH)}_2\)) release hydroxide ions in a similar way strong acids release hydronium ions. This means each unit of these bases will dissociate completely to release hydroxide ions.
For KOH with a concentration of 0.0087 M:
- Since KOH is a strong base, it fully dissociates, resulting in 0.0087 M of hydroxide ions.
The presence of these ions dictates the basicity of the solutions and influences acid-base equilibria.
Strong Acids and Bases
Strong acids and bases are substances that completely dissociate into their constituent ions in a solution. Because of this complete dissociation, the concentration of ions generated equals the initial concentration of the acid or base, simplifying the calculation of ion concentrations.
Strong acids, exemplified by HI and HNO₃, dissociate to form hydronium ions. This increase in hydronium ions results in a lower pH (more acidic). For example, a 5.8 x 10⁻⁴ M solution of HI, a strong acid, directly corresponds to a hydronium ion concentration of 5.8 x 10⁻⁴ M.
Similarly, strong bases like KOH and \(\text{Sr(OH)}_2\) dissociate to produce hydroxide ions.
Strong acids, exemplified by HI and HNO₃, dissociate to form hydronium ions. This increase in hydronium ions results in a lower pH (more acidic). For example, a 5.8 x 10⁻⁴ M solution of HI, a strong acid, directly corresponds to a hydronium ion concentration of 5.8 x 10⁻⁴ M.
Similarly, strong bases like KOH and \(\text{Sr(OH)}_2\) dissociate to produce hydroxide ions.
- In the case of 0.0087 M KOH, this leads to 0.0087 M hydroxide ions.
- For \(\text{Sr(OH)}_2\), since each molecule provides double the hydroxide ions, a concentration of 0.00213 M produces 0.00426 M of hydroxide ions.
Other exercises in this chapter
Problem 4
Which of the following species are amphiprotic in aqueous solution? For such a species, write one equation showing it acting as an acid, and another equation sh
View solution Problem 5
With which of the following bases will the ionization of acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH},\) proceed furthest toward completion (to the right): (a)
View solution Problem 10
What is the pH of each of the following solutions? (a) \(0.0045 \mathrm{M} \mathrm{HCl} ;\) (b) \(6.14 \times 10^{-4} \mathrm{M} \mathrm{HNO}_{3} ;\) (c) 0.0068
View solution Problem 11
Calculate \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\mathrm{pH}\) in saturated \(\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})\) which contains \(3.9 \ma
View solution