Calculate the pH of a solution that is 0.0330 M in a. NaH,PO4 (Kal = 7.11 x 10-3, K,2 = 6.32 × 10-8). %3D pH = b. NaHSO3 (Kal = 1.23 x 10-2, K2 = 6.60 × 10-8). pH = H2NC2 H4NH3+Cl¯ (Kal = 1.42 × 10-7, K.2 с. = 1.18 x 10-10). pH =

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Chapter1: Chemical Foundations
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**Calculate the pH of a solution that is 0.0330 M in:**

a. **NaH₂PO₄** \((K_{a1} = 7.11 \times 10^{-3}, \ K_{a2} = 6.32 \times 10^{-8})\).

   pH = \_\_\_

b. **NaHSO₃** \((K_{a1} = 1.23 \times 10^{-2}, \ K_{a2} = 6.60 \times 10^{-8})\).

   pH = \_\_\_

c. **H₂NC₂H₄NH₃⁺Cl⁻** \((K_{a1} = 1.42 \times 10^{-7}, \ K_{a2} = 1.18 \times 10^{-10})\).

   pH = \_\_\_

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**Instructions for Calculating pH:**

- Determine the dominant acidic or basic species in the solution.
- Use the appropriate equilibrium expression and initial concentrations to calculate the concentration of hydrogen ions \([H⁺]\) or hydroxide ions \([OH⁻]\).
- Convert the concentration to pH using:
  - \( \text{pH} = -\log[H⁺] \)
- For solutions with multiple dissociation steps, you may need to consider each equilibrium step to determine the final pH accurately.
Transcribed Image Text:**Calculate the pH of a solution that is 0.0330 M in:** a. **NaH₂PO₄** \((K_{a1} = 7.11 \times 10^{-3}, \ K_{a2} = 6.32 \times 10^{-8})\). pH = \_\_\_ b. **NaHSO₃** \((K_{a1} = 1.23 \times 10^{-2}, \ K_{a2} = 6.60 \times 10^{-8})\). pH = \_\_\_ c. **H₂NC₂H₄NH₃⁺Cl⁻** \((K_{a1} = 1.42 \times 10^{-7}, \ K_{a2} = 1.18 \times 10^{-10})\). pH = \_\_\_ --- **Instructions for Calculating pH:** - Determine the dominant acidic or basic species in the solution. - Use the appropriate equilibrium expression and initial concentrations to calculate the concentration of hydrogen ions \([H⁺]\) or hydroxide ions \([OH⁻]\). - Convert the concentration to pH using: - \( \text{pH} = -\log[H⁺] \) - For solutions with multiple dissociation steps, you may need to consider each equilibrium step to determine the final pH accurately.
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