**Problem**: Calculate the pH of a 0.025 M HCl solution.**Solution**: Since HCl is a strong acid, it will completely dissociate in water. The concentration of H⁺ ions will be 0.025 M. So, pH = -log[H⁺] = -log(0.025) = 1.60.**Problem**: What is the pH of a 0.010 M NaOH solution?**Solution**: NaOH is a strong base and will completely dissociate in water. The concentration of OH⁻ ions will be 0.010 M. We first calculate pOH = -log[OH⁻] = -log(0.010) = 2. Then, use the formula pH = 14 – pOH = 14 – 2 = 12.**Problem**: Determine the pH of a 0.100 M HNO₃ solution.**Solution**: HNO₃ is a strong acid, it will dissociate completely in water. The concentration of H⁺ ions will be 0.100 M. So, pH = -log[H⁺] = -log(0.100) = 1.**Problem**: What is the pH of a 0.075 M KOH solution?**Solution**: KOH is a strong base and will completely dissociate in water. The concentration of OH⁻ ions will be 0.075 M. We first calculate pOH = -log[OH⁻] = -log(0.075) = 1.12. Then, use the formula pH = 14 – pOH = 14 – 1.12 = 12.88.**Problem**: Calculate the pH of a solution with [H⁺] = 1 x 10⁻⁴ M.**Solution**: pH = -log[H⁺] = -log(1 x 10⁻⁴) = 4.**Problem**: What is the pH of a solution with [OH⁻] = 1 x 10⁻⁵ M?**Solution**: First, calculate the pOH = -log[OH⁻] = -log(1 x 10⁻⁵) = 5. Then, pH = 14 – pOH = 14 – 5 = 9.**Problem**: Determine the pH of a 0.002 M H₂SO₄ solution.**Solution**: H₂SO₄ is a strong acid, it will dissociate completely in water. The concentration of H⁺ ions will be 2 x 0.002 M = 0.004 M. So, pH = -log[H⁺] = -log(0.004) = 2.40.- Problem: Determine the pH of a 0.1 M solution of hydrochloric acid (HCl). Solution: HCl is a strong acid, which means it completely ionizes in water. Therefore, the concentration of hydronium ions [H₃O⁺] equals the concentration of HCl. The pH is calculated as pH = -log[H₃O⁺] = -log[0.1] = 1.
- Problem: Calculate the pOH of a 0.01 M solution of NaOH. Solution: NaOH is a strong base, and completely ionizes to OH⁻ ions. pOH = -log[OH⁻] = -log[0.01] = 2.
- Problem: Determine the pH of a solution with [OH⁻] = 1×10⁻⁴ M. Solution: First calculate the pOH using pOH = -log[OH⁻] = -log[1×10⁻⁴] = 4. Then, use the relationship between pH and pOH in water at 25°C, which is pH + pOH = 14, to find pH = 14 – 4 = 10.
- Problem: What is the [H₃O⁺] in a solution with a pH of 3? Solution: Use the definition of pH. [H₃O⁺] = 10⁻pH = 10⁻³ M.
- Problem: What is the [OH⁻] in a solution with a pOH of 5? Solution: Use the definition of pOH. [OH⁻] = 10⁻pOH = 10⁻⁵ M.
- Problem: Calculate the pH of a solution where [H₃O⁺] = 1×10⁻⁶ M. Solution: Use the definition of pH: pH = -log[H₃O⁺] = -log[1×10⁻⁶] = 6.
- Problem: Calculate the pOH of a solution where [OH⁻] = 1×10⁻⁷ M. Solution: Use the definition of pOH: pOH = -log[OH⁻] = -log[1×10⁻⁷] = 7.
- Problem: If the [OH⁻] in a solution is 1×10⁻⁶ M, what is the pH? Solution: First calculate the pOH = -log[1×10⁻⁶] = 6. Then, using the relationship between pH and pOH, find that the pH = 14 – pOH = 14 – 6 = 8.
- Problem: If the pH of a solution is 5, what is the [OH⁻]? Solution: First calculate the pOH = 14 – pH = 14 – 5 = 9. Then, find the [OH⁻] using [OH⁻] = 10⁻pOH = 10⁻⁹ M.
- Problem: Calculate the [H₃O⁺] in a solution with a pOH of 2. Solution: First find the pH = 14 – pOH = 14 – 2 = 12. Then, find the [H₃O⁺] using [H₃O⁺] = 10⁻pH = 10⁻¹² M.
- Problem: Determine the pH of a solution with a [H₃O⁺] of 1×10⁻⁴ M. Solution: Use the definition of pH: pH = -log[H₃O⁺] = -log[1×10⁻⁴] = 4.
Problem: Determine the pOH of a solution with a [OH⁻] of 1×10⁻³ M. Solution: Use the definition of pOH: pOH = -log[OH⁻] = -log[1×10⁻³] = 3.