--- module: 105-02 language: en chapter: 105 title: "Quantitative Chemistry and Acid–Base Foundations" module_title: "Acid–base equilibria, buffers, and titration reasoning" source_sha256: ceeb6d28a81a4bba446c7612d656837980d2107618bb0bb19b7771d882f664f8 --- # Acid–base equilibria, buffers, titration ## Framing the problem ### Identify proton donor, acceptor, conserved quantities ### Near-complete reaction or reversible equilibrium #### Confusing the stages wrecks buffer calculations ### Ideal dilute aqueous solutions at 25 °C ## Acid and base definitions ### Brønsted–Lowry: acid donates, base accepts proton ### Conjugate pair differs by one proton and one charge ### Lewis: acid accepts, base donates electron pair #### Proton transfer fits within Lewis view #### Lewis interaction need not transfer a proton ### State which definition is in use ## Logarithmic measurement ### pH: negative log of hydrogen-ion activity #### Dilute problems approximate activity by concentration ### One pH unit is a tenfold activity change #### Two units is a hundredfold ratio #### Averaging pH values misdescribes mixtures ### Water's ion product about 10^-14 at 25 °C #### pH plus pOH about 14 ### Neutral means equal H+ and OH- activities #### Neutral pH shifts with the ion product ### 0 to 14 is not an absolute boundary ## Strength versus concentration ### Ka compares product activities with acid activity #### Larger Ka, smaller pKa: stronger acid ### Concentration is acid supplied per volume ### Dilute strong vs concentrated weak: words cannot rank ### Worked weak acid: 0.040 M, Ka 10^-5 #### Small-change approximation gives pH about 3.20 #### Only about 1.58% dissociates #### Exact quadratic gives nearly the same pH ### Check assumptions after calculating #### Large dissociation: keep depleted acid term #### 10^-8 M strong acid does not give pH 8 #### Water equilibrium puts pH slightly below 7 ## Buffers as finite reserves ### Weak acid with conjugate base, or weak-base pair ### Added acid consumes base component, and vice versa #### pH changes less than unbuffered solution ### Henderson–Hasselbalch: pH = pKa + log(base/acid) #### Neither component may be zero ### Equal 10.0 mmol pair at pKa 5 gives pH 5 #### 2.00 mmol strong acid converts A- to HA #### Post-reaction ratio 8:12 gives pH 4.82 #### Initial ratio wrongly predicts no change ## Buffer capacity ### Same pH with only 1.00 mmol each #### Base exhausted, excess strong acid remains #### No longer a two-component buffer ### Capacity depends on amount, not pH alone ### Dilution keeps ratio and roughly pH #### Fixed volume then holds less capacity ## Titration regions ### Equivalence point: stoichiometric amounts reacted ### Endpoint: indicator or instrument signal #### Close in a good method, not the same event ### Before equivalence: acid and conjugate base coexist ### Half-equivalence: pH approximates pKa ### Example: 2.00 mmol weak acid, 0.100 M base #### Equivalence at 20.0 mL, half at 10.0 mL ### Beyond equivalence, excess strong base dominates #### 25.0 mL leaves 0.50 mmol in 50.0 mL #### OH- 0.0100 M, pH 12 ### Each region needs a different model ## Equivalence-point pH ### Conjugate base reacts with water #### Weak acid, strong base: pH above 7 #### Strong acid, strong base: neutral, no hydrolysis ### Ka times Kb equals water ion product #### Ka 10^-5 gives Kb 10^-9 ### 2.00 mmol in 45.0 mL is about 0.0444 M ### OH- about square root of Kb times concentration #### pOH about 5.18, pH about 8.82 ## Mixing strong acids ### Equal volumes at pH 2 and pH 4 #### H+ about 0.00505 M, pH about 2.30, not 3 #### Concentrated solution supplies most H+ ### Assumes additive volumes, negligible water ## Protonation in biology ### One unit above pKa: about 10 deprotonated per 1 ### One unit below: the reverse ratio ### Proteins have multiple interacting groups #### One group cannot fix whole-protein charge ### Polyprotic acids lose protons stepwise ### Solving habit #### Identify species, finish dominant stoichiometry #### Choose equilibrium model #### Test charge balance and direction of change