CHE 141 1st Edition Lecture 18Outline- Buffer Solutions- Calcuating pH Changes in a Buffer Solution- Buffer Capacity and RangeBuffer Solutions- Buffer a solution which resists pH changes upon addition of an acid or a base- Buffers contain both an acidic species to neutralize OH- ions and a basicone to neutralize H3O+ ions- Buffers are often prepared by mixing a weak acid or base with a salt of that acid or base- The acidic and basic species of the buffer must not wholly consume each other viaa neutralization reaction- This is why strong acid/bases don’t work- Consider a buffer solution of HF and its salt NaF- In aqueous solution, the weak acid partially dissociates: HF(aq)+H2O(l) reversible F- +H3O+- NaF dissolves completely in water: NaFNa+(aq)+F-(aq)- By Le Chateliers principle the common ion F- will inhibit diccociation of HFThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- If acid is added to this buffer solution, the F- base (from the salt) will neutralize added acid: F-(aq)+H3O+(aq) reversible HF(aq)+H2O(l) - The equilibrium for this reaction lies far to the right, decreasing [H3O+] and increasing [HF]- If base is added to this buffer solution, the HF will neutralize added base: HF(aq)+OH-(l) reversible F-+H2O(l)- This equilibrium lies to the right, decreasing [OH]- and increasing [F-]- In general, for any buffer- If H3O+ ions are added, they react with the base component of the buffer: H3O+(aq)+A-(aq)HA(aq)+H2O(l)- Adding acid creates more weak acid- If OH- ions are added to the buffered solution they react with the acid component of the buffer: OH-(aq)+HA(aq)H2O(l)+A-(aq)- Adding base creates more conjugate base- Remember when we derive the Henderson Hasselbalch equation we rearranged the generic Ka expression in terms of H3O+: H3O+=Ka[HA]/[A-]- Buffers resist pH changes best when [HA] is about equal to [A-] which corresponds to [H3O+]=Ka- We usually use a buffer with pKa close to desired pH we want to maintainCalcuating pH Changes in a Buffer Solution- Although buffers resist pH changes, addition of a acid or a base to a buffer does still result in a small change in pH- To calculate this we need to break the problem up into two parts: o The stoichiometry Calculation: in which we calculate how the addition changes the relative amounts of acid and conjugate baseo The equilibrium calculation: in which we calculate the pH based on the new amounts of acid and conjugate base- Note that for the equilibrium step, we could also have used the Henderson Hasselbalch equation (still need to do stoichiometry calculation)- Because this expression contains ration of concentrations, we can use the amounts of acid and base in terms of moles instead of concentration (the volume is the same for both the acid and base, thus the volumes cancel out)Buffer Capacity and Range- Buffer capacity is the quantity of acid or base that a buffer can neutralize while maintaining its pH within a desired range- A buffer with equal amounts of acid and conjugate base is more resistant to pH change: pH=pKa+log[base]/[acid]- If [base]=[acid], then pH=pKa- The relative concentration of acid and conjugate base should not differ by more than afactor of 10- A buffer is most effective when the concentrations of acid and conjugate base are high- Buffer range is the pH range in which a given buffer can provide pH protection- pH=pKa+log[base]/[acid]- lowest pH for effective buffer occurs when base is one tenth as concentration as the acid: pH=pKa+log.10=pKa-1- Highest pH for effective buffer occurs when the base is 10 times as concentrated as the acid: pH=pKa+log10=pKa+1- Buffer range=pKa plus or minus 1- We can rearrange the Henderson Hasselbalch equation in terms of the ratio of base to acid: pH=pKa+log[base]/[acid], log[base]/[acid]=pH-pKa,
View Full Document
Unlocking...