Here are the solutions for Exercise 3.10: 1. Which solution has a higher $[\text{H}_3\text{O}^+]$ concentration, one with $\text{pH}=5$, or with $\text{pH}=8$? Step 1: Calculate $[\text{H}_3\text{O}^+]$ for each $\text{pH}$ value using the formula $[\text{H}_3\text{O}^+] = 10^{-\text{pH}}$. For $\text{pH}=5$: $$[\text{H}_3\text{O}^+] = 10^{-5} \text{ M}$$ For $\text{pH}=8$: $$[\text{H}_3\text{O}^+] = 10^{-8} \text{ M}$$ Step 2: Compare the concentrations. Since $10^{-5} > 10^{-8}$, the solution with $\text{pH}=5$ has a higher $[\text{H}_3\text{O}^+]$ concentration. Answer: The solution with $\text{pH}=5$ has a higher $[\text{H}_3\text{O}^+]$ concentration. 2. Which solution has the higher $\text{OH}^-$ concentration? Step 1: Calculate $\text{pOH}$ for each $\text{pH}$ value using the formula $\text{pOH} = 14 - \text{pH}$. For $\text{pH}=5$: $$\text{pOH} = 14 - 5 = 9$$ For $\text{pH}=8$: $$\text{pOH} = 14 - 8 = 6$$ Step 2: Calculate $[\text{OH}^-]$ for each $\text{pOH}$ value using the formula $[\text{OH}^-] = 10^{-\text{pOH}}$. For $\text{pOH}=9$: $$[\text{OH}^-] = 10^{-9} \text{ M}$$ For $\text{pOH}=6$: $$[\text{OH}^-] = 10^{-6} \text{ M}$$ Step 3: Compare the concentrations. Since $10^{-6} > 10^{-9}$, the solution with $\text{pH}=8$ has a higher $[\text{OH}^-]$ concentration. Answer: The solution with $\text{pH}=8$ has a higher $[\text{OH}^-]$ concentration. 3. Give the $\text{pH}$ and $\text{pOH}$ for each solution with the following concentrations: a) $[\text{H}_3\text{O}^+] = 1.0 \times 10^{-3}$ Step 1: Calculate $\text{pH}$ using the formula $\text{pH} = -\log[\text{H}_3\text{O}^+]$. $$\text{pH} = -\log(1.0 \times 10^{-3}) = 3.0$$ Step 2: Calculate $\text{pOH}$ using the formula $\text{pOH} = 14 - \text{pH}$. $$\text{pOH} = 14 - 3.0 = 11.0$$ Answer: $\boxed{\text{pH} = 3.0, \text{pOH} = 11.0}$ b) $[\text{OH}^-] = 1.0 \times 10^{-9} \text{ M}$ Step 1: Calculate $\text{pOH}$ using the formula $\text{pOH} = -\log[\text{OH}^-]$. $$\text{pOH} = -\log(1.0 \times 10^{-9}) = 9.0$$ Step 2: Calculate $\text{pH}$ using the formula $\text{pH} = 14 - \text{pOH}$. $$\text{pH} = 14 - 9.0 = 5.0$$ Answer: $\boxed{\text{pH} = 5.0, \text{pOH} = 9.0}$ 4. Give the hydroxide ion concentrations of solutions with the following values of $\text{pH}$: a) $\text{pH } 13.0$ Step 1: Calculate $\text{pOH}$ using the formula $\text{pOH} = 14 - \text{pH}$. $$\text{pOH} = 14 - 13.0 = 1.0$$ Step 2: Calculate $[\text{OH}^-]$ using the formula $[\text{OH}^-] = 10^{-\text{pOH}}$. $$[\text{OH}^-] = 10^{-1.0} \text{ M} = 0.1 \text{ M}$$ Answer: $\boxed{[\text{OH}^-] = 0.1 \text{ M}}$ b) $\text{pH } 3.0$ Step 1: Calculate $\text{pOH}$ using the formula $\text{pOH} = 14 - \text{pH}$. $$\text{pOH} = 14 - 3.0 = 11.0$$ Step 2: Calculate $[\text{OH}^-]$ using the formula $[\text{OH}^-] = 10^{-\text{pOH}}$. $$[\text{OH}^-] = 10^{-11.0} \text{ M}$$ Answer: $\boxed{[\text{OH}^-] = 10^{-11.0} \text{ M}}$ c) $\text{pH } 8.0$ Step 1: Calculate $\text{pOH}$ using the formula $\text{pOH} = 14 - \text{pH}$. $$\text{pOH} = 14 - 8.0 = 6.0$$ Step 2: Calculate $[\text{OH}^-]$ using the formula $[\text{OH}^-] = 10^{-\text{pOH}}$. $$[\text{OH}^-] = 10^{-6.0} \text{ M}$$ Answer: $\boxed{[\text{OH}^-] = 10^{-6.0} \text{ M}}$ 5. Which of the solutions is most acidic? Which is most basic? (Referring to the solutions from part 4: $\text{pH } 13.0$, $\text{pH } 3.0$, $\text{pH } 8.0$) • Acidity increases as $\text{pH}$ decreases. The lowest $\text{pH}$ among $13.0$, $3.0$, and $8.0$ is $3.0$. • Basicity increases as $\text{pH}$ increases. The highest $\text{pH}$ among $13.0$, $3.0$, and $8.0$ is $13.0$. Answer: The solution with $\text{pH}=3.0$ is most acidic. The solution with $\text{pH}=13.0$ is most basic.