1. Write the equation and prepare an ICE chart.
............NH3 + HOH ==> NH4^+ + OH^-
initial......0.100........0........0
change.......-x..........+x.......+x
final.......0.100-x.......x.........x
Kb = (NH4^+)(OH^-)/(NH3)
Substitute into Kb expression from the ICE chart above.
Kb = 1.8E-5 from tables. It will be in your book or in your notes.
1.8E-5 = (x)(x)/0.1-x
Solve for x = OH^- in this case, convert to pOH, then to pH.
The second one is done the same way but it is an acid and not a base. Oxalic has has two Ka values but the problem tells you to ignore the second one and use only k1.
Post your work if you get stuck.
Determine the pH of the following, giving your answers to two decimal places. I don't know how to do these problems =(
a. 0.100 M ammonia
b. 0.11 M oxalic acid. Assume that dissociation resulting from Ka2 is negligible relative to that from Ka1
3 answers
1. not sure where to find Kb as listed as 1.8E^-5?
so is it 1.8E^-5= (NH4+)(NH3)/0.1-X
(9.25)(4.75)/0.1-X = 439.375-X
and how to I convert this to pOH and then to pH?
so is it 1.8E^-5= (NH4+)(NH3)/0.1-X
(9.25)(4.75)/0.1-X = 439.375-X
and how to I convert this to pOH and then to pH?
No. I gave you the formula for finding Kb. They aren't listed in tables because you can calculate them. Kb = (Kw/Ka) = (1E-14/1.8E-5) = 5.55E-10
2 answers