Question
A 0.035 M solution of a weak base has a pH of 10.88. What is the Kb for the acid?
Where Di I even begun**???
Where Di I even begun**???
Answers
DrBob222
You ALWAYS start with one or all of the following:
1. balanced equation
2. ICE chart
3. think in terms of mols.
pH = 10.88 so pOH = 3.12 and OH^- = 7.58E-4
M..........BOH ==> B^+ + OH^-
initial.0.035M.....0......0
change..7.58E-4..7.58E-4..7.58E-4
equil..0.035-7.58E-4....etc
Kb = (B^+)(OH^-)/(BOH)
Substitute from the ICE chart and solve for Kb. Note that 0.035-7.58E-4 is not = 0.035
1. balanced equation
2. ICE chart
3. think in terms of mols.
pH = 10.88 so pOH = 3.12 and OH^- = 7.58E-4
M..........BOH ==> B^+ + OH^-
initial.0.035M.....0......0
change..7.58E-4..7.58E-4..7.58E-4
equil..0.035-7.58E-4....etc
Kb = (B^+)(OH^-)/(BOH)
Substitute from the ICE chart and solve for Kb. Note that 0.035-7.58E-4 is not = 0.035