Summary of "17.3a Strong Acid Strong Base Titrations pH Calculations | General Chemistry"
Summary of “17.3a Strong Acid Strong Base Titrations pH Calculations | General Chemistry”
This lesson, presented by Chad from Chad’s Prep, focuses on pH calculations during strong acid-strong base titrations. The goal is to provide a clear understanding of the process rather than rote memorization, helping students grasp why certain calculation steps are performed.
Main Ideas and Concepts
- Strong acid-strong base titrations involve the neutralization reaction between a strong acid (e.g., HCl) and a strong base (e.g., NaOH) that goes to completion.
- The pH changes during the titration can be divided into three key regions:
- Before the equivalence point: excess strong acid remains, so the solution is acidic (pH < 7).
- At the equivalence point: moles of acid equal moles of base, resulting in a neutral solution (pH = 7).
- After the equivalence point: excess strong base remains, so the solution is basic (pH > 7).
Methodology for pH Calculations in Strong Acid-Strong Base Titrations
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Initial point (no base added):
- Calculate pH by taking the negative logarithm of the strong acid concentration.
- Example: For 0.1 M HCl, [ \text{pH} = -\log(0.1) = 1 ]
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Before the equivalence point (excess acid present):
- Calculate moles of acid and base: [ \text{moles} = \text{molarity} \times \text{volume (L)} ]
- Determine the limiting reagent (usually the base if less added).
- Calculate excess moles of acid remaining: [ \text{excess acid moles} = \text{initial acid moles} - \text{moles of base added} ]
- Calculate concentration of excess acid: [ [\text{H}^+] = \frac{\text{excess moles of acid}}{\text{total volume (L)}} ]
- Calculate pH: [ \text{pH} = -\log[\text{H}^+] ]
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At the equivalence point:
- Moles of acid = moles of base.
- All acid and base neutralize completely.
- No excess H⁺ or OH⁻ remain, so [ \text{pH} = 7 ]
- No further calculation needed beyond confirming mole equality.
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After the equivalence point (excess base present):
- Calculate moles of acid and base as before.
- Determine excess moles of base remaining: [ \text{excess base moles} = \text{moles of base added} - \text{initial acid moles} ]
- Calculate concentration of excess hydroxide: [ [\text{OH}^-] = \frac{\text{excess moles of base}}{\text{total volume (L)}} ]
- Calculate pOH: [ \text{pOH} = -\log[\text{OH}^-] ]
- Calculate pH: [ \text{pH} = 14 - \text{pOH} ]
Additional Notes
- The reaction between strong acid and strong base is a limiting reagent problem because it goes to completion.
- Neutral salts like NaCl formed during the reaction do not affect pH.
- Volumes are additive when mixing solutions (e.g., 150 mL acid + 100 mL base = 250 mL total volume).
- Concentrations must be recalculated after mixing due to volume changes.
- The lesson emphasizes understanding the underlying chemistry rather than just memorizing formulas.
- Future lessons will cover more complex cases such as weak acid-strong base titrations where salt effects and buffer regions must be considered.
Summary of Calculation Steps
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Calculate moles of acid and base: [ \text{moles} = \text{molarity} \times \text{volume (L)} ]
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Determine limiting reagent and excess moles:
- Before equivalence: [ \text{excess acid moles} = \text{acid moles} - \text{base moles} ]
- After equivalence: [ \text{excess base moles} = \text{base moles} - \text{acid moles} ]
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Calculate concentration of excess ion: [ \text{concentration} = \frac{\text{excess moles}}{\text{total volume (L)}} ]
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Calculate pH or pOH:
- For excess acid: [ \text{pH} = -\log[\text{H}^+] ]
- For excess base: [ \text{pOH} = -\log[\text{OH}^-] ] [ \text{pH} = 14 - \text{pOH} ]
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At equivalence point: [ \text{pH} = 7 \quad (\text{neutral}) ]
Speaker / Source
- Chad — Instructor and creator of Chad’s Prep YouTube channel and general chemistry lessons.
Category
Educational
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