HA(aq) + OH- --> A-(aq) + H2O(L)
if weak acid is in excess --> a buffer --> Hendeson Hasselbalch equaiton
if strong base is in excess, the moles of excess hydroxide ions is used for pH
if acid and base are equimolar then the equilibrium concentr
strong acid strong base equivalence point
pH = 7 @ 25 C / 298 K
how to increase/decrease buffering capacity
equivalence point
H2O pH is always...
7, could be acid (pH) or base (pOH)
1.0 x 10^-14 at 25 C or 298 K
14 always!!
how do you calculate a buffer's pH
pH = pKa + log [A-]/[HA] ---> acids
pOH = pKb + log [BH+]/[B] ---> bases
the conjugate salt is on top, and the bottom is either the base or the acid
assume there is a limiting and excess reagent (because it will contain either a strong acid or base)
pH = pKa
strong acid-weak base reaction
if weak base is in excess --> a buffer --> Hendeson Hasselbalch equaiton
if strong acid is in excess, the moles of excess acid are used to determine pH
if acid and base are equimolar then the equilibrium concentrations are used to determine pH
what is a buffer
in what range are buffers effective
within 1 pH of its pKa, up or down
