pKa pKb pOH other than pH
! sms mms setting timezone and clock face
pH = pKa@half-equivalence-point "mL" milliliter pronounced like 'mills'
"HA" read 'acid'
"A${}^-$" read conjugate 'base'
cheat sheet for case: weak acid - strong base titration
V > Veq ... strong base $\frac{extra strong base (mol)}{total vol} = $[OH${}^-$] pOH->pH
V == Veq ... (all acid now gone(!salt?). weak (conjugate) base problem) A${}^-$ + H${}_2$O⇔HA+OH${}^-$ found [OH${}^-$]
V == Vhalf ... pH = pKa
0 < V < Veq [buffering region] use Henderson-Hasselbalch Eq. iff 5% rule satisfies pH = pKa - $log\frac{[HA]}{[A^-]}$
V == 0 weadk acid ... HA + H${}_2$O ⇔ H${}_3$O + A${}^-$ use Ka to find [H${}_3$O]
disproportionation reaction for redax ... like amphoric for acid-base
Lec 24 | MIT 5.111 Principles of Chemical Science, Fall 2005
! sms mms setting timezone and clock face
pH = pKa@half-equivalence-point "mL" milliliter pronounced like 'mills'
"HA" read 'acid'
"A${}^-$" read conjugate 'base'
cheat sheet for case: weak acid - strong base titration
V > Veq ... strong base $\frac{extra strong base (mol)}{total vol} = $[OH${}^-$] pOH->pH
V == Veq ... (all acid now gone(!salt?). weak (conjugate) base problem) A${}^-$ + H${}_2$O⇔HA+OH${}^-$ found [OH${}^-$]
V == Vhalf ... pH = pKa
0 < V < Veq [buffering region] use Henderson-Hasselbalch Eq. iff 5% rule satisfies pH = pKa - $log\frac{[HA]}{[A^-]}$
V == 0 weadk acid ... HA + H${}_2$O ⇔ H${}_3$O + A${}^-$ use Ka to find [H${}_3$O]
disproportionation reaction for redax ... like amphoric for acid-base
Lec 24 | MIT 5.111 Principles of Chemical Science, Fall 2005
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