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A buffer solution is made up of 1.0 M NH3 and 1.0 M NH4CI. = 1.8 x 10-5 K₂NH3 What is the pH when 20 mL of 0.60 M NaOH is added to 25. mL of the buffer? pH after base added [?] pH Enter

A buffer solution is made up of 1.0 M NH3 and 1.0 M NH4CI. = 1.8 x 10-5 K₂NH3 What is the pH when 20 mL of 0.60 M NaOH is added to 25. mL of the buffer? pH after base added [?] pH Enter

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**Buffer Solution pH Calculation** A buffer solution is composed of 1.0 M ammonia (NH₃) and 1.0 M ammonium chloride (NH₄Cl). Given: - \( K_b \) for NH₃ = \( 1.8 \times 10^{-5} \) **Problem Statement:** What is the pH when 20 mL of 0.60 M sodium hydroxide (NaOH) is added to 25 mL of the buffer solution? The equation to determine the new pH is: \[ \text{pH after base added} = \text{[?]} \] ### Steps to Solve: 1. **Determine the moles of NH₄⁺ and NH₃ initially in the buffer:** - Volume of buffer = 25 mL → 0.025 L - Molarity of NH₄Cl = 1.0 M - Moles of NH₄⁺ = 0.025 L * 1.0 M = 0.025 moles - Moles of NH₃ = 0.025 L * 1.0 M = 0.025 moles 2. **Determine the moles of NaOH added:** - Volume of NaOH = 20 mL → 0.020 L - Molarity of NaOH = 0.60 M - Moles of NaOH = 0.020 L * 0.60 M = 0.012 moles 3. **Calculate the changes in moles after NaOH addition:** - NaOH reacts with NH₄⁺ to form NH₃: \[ \text{NH₄⁺} + \text{OH⁻} → \text{NH₃} + \text{H₂O} \] - Moles of NH₄⁺ after reaction = 0.025 moles - 0.012 moles = 0.013 moles - Moles of NH₃ after reaction = 0.025 moles + 0.012 moles = 0.037 moles 4. **Use the Henderson-Hasselbalch equation to find the new pH:** \[ \text{pH} = \text{pKa}
Transcribed Image Text:**Buffer Solution pH Calculation** A buffer solution is composed of 1.0 M ammonia (NH₃) and 1.0 M ammonium chloride (NH₄Cl). Given: - \( K_b \) for NH₃ = \( 1.8 \times 10^{-5} \) **Problem Statement:** What is the pH when 20 mL of 0.60 M sodium hydroxide (NaOH) is added to 25 mL of the buffer solution? The equation to determine the new pH is: \[ \text{pH after base added} = \text{[?]} \] ### Steps to Solve: 1. **Determine the moles of NH₄⁺ and NH₃ initially in the buffer:** - Volume of buffer = 25 mL → 0.025 L - Molarity of NH₄Cl = 1.0 M - Moles of NH₄⁺ = 0.025 L * 1.0 M = 0.025 moles - Moles of NH₃ = 0.025 L * 1.0 M = 0.025 moles 2. **Determine the moles of NaOH added:** - Volume of NaOH = 20 mL → 0.020 L - Molarity of NaOH = 0.60 M - Moles of NaOH = 0.020 L * 0.60 M = 0.012 moles 3. **Calculate the changes in moles after NaOH addition:** - NaOH reacts with NH₄⁺ to form NH₃: \[ \text{NH₄⁺} + \text{OH⁻} → \text{NH₃} + \text{H₂O} \] - Moles of NH₄⁺ after reaction = 0.025 moles - 0.012 moles = 0.013 moles - Moles of NH₃ after reaction = 0.025 moles + 0.012 moles = 0.037 moles 4. **Use the Henderson-Hasselbalch equation to find the new pH:** \[ \text{pH} = \text{pKa}
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