General Chemistry 2
Chemistry Book Notes:
Chapter 21: Buffers and the Titration of Acids and Bases
21-1 Henderson-Hasselbalch Equation
THE HH EQUATION OFTEN CAN BE USED TO CALCULATE THE pH OF A BUFFER SOLUTION
-buffer >a solution containing both a weak acid and its conjugate base can resist a change in pH by neutralizing either an added acid or an added base. Ex. acetic acid-acetate soln (acid with conj. Base) > Kc for a buffer reaction can = 1/Ka or 1/Kb if you add an acid or base because it is the reverse of an acid or base equlibrium reaction **the larger the Kc value indicates that the reaction goes essentially to completion and that all the acid/base added converts to the acid/base in the soln ex. in this example H3O+ ( CH3COOH; OH- ( CH3COO-
-HH equation > pH= pKa + log (base/acid) (initial concentrations) *Conditions for HH eqn: 1) value of Ka for the conjugate acid-base pair should be in the range of 10^-4 -10^-11 2) ratio of base and acid must be between .1 and 10 3) values of acid and base should be between 10^-3 and 1 M
21-2 Buffers
BUFFER SOLUNTIONS SUPRESS A CHANGE IN pH WHEN A SMALL AMT OF EITHER AN ACID OR BASE IS ADDED
-when working with dilute solutions we work in mmol
-calculating mmols of base and acid when an more acid is added (continue with CH3COO example) >mmol of CH3COO- after HCl added= (mmol of CH3COO-before HCl added) – (mmol of HCl added) ** + for CH3COOH because you are adding an acid
-buffer capacity: the capacity of a buffer to resist changes in pH is not unlimited >if you add enough acid to neutralize the conjugate base then the pH will drop more >if you add enough base to neutralize the conjugate acid then the pH will rise more
-buffer dilution >buffers resist change even when they are diluted >if you dilute a buffer by a factor of two then the concentrations of the acid and base are decreased by a factor of two but their ratio is still constant **therefore pH does not
Chapter 21: Buffers and the Titration of Acids and Bases
21-1 Henderson-Hasselbalch Equation
THE HH EQUATION OFTEN CAN BE USED TO CALCULATE THE pH OF A BUFFER SOLUTION
-buffer >a solution containing both a weak acid and its conjugate base can resist a change in pH by neutralizing either an added acid or an added base. Ex. acetic acid-acetate soln (acid with conj. Base) > Kc for a buffer reaction can = 1/Ka or 1/Kb if you add an acid or base because it is the reverse of an acid or base equlibrium reaction **the larger the Kc value indicates that the reaction goes essentially to completion and that all the acid/base added converts to the acid/base in the soln ex. in this example H3O+ ( CH3COOH; OH- ( CH3COO-
-HH equation > pH= pKa + log (base/acid) (initial concentrations) *Conditions for HH eqn: 1) value of Ka for the conjugate acid-base pair should be in the range of 10^-4 -10^-11 2) ratio of base and acid must be between .1 and 10 3) values of acid and base should be between 10^-3 and 1 M
21-2 Buffers
BUFFER SOLUNTIONS SUPRESS A CHANGE IN pH WHEN A SMALL AMT OF EITHER AN ACID OR BASE IS ADDED
-when working with dilute solutions we work in mmol
-calculating mmols of base and acid when an more acid is added (continue with CH3COO example) >mmol of CH3COO- after HCl added= (mmol of CH3COO-before HCl added) – (mmol of HCl added) ** + for CH3COOH because you are adding an acid
-buffer capacity: the capacity of a buffer to resist changes in pH is not unlimited >if you add enough acid to neutralize the conjugate base then the pH will drop more >if you add enough base to neutralize the conjugate acid then the pH will rise more
-buffer dilution >buffers resist change even when they are diluted >if you dilute a buffer by a factor of two then the concentrations of the acid and base are decreased by a factor of two but their ratio is still constant **therefore pH does not