a). NH3, NH4^+, OH^-, H3O^+
b) NH3 is a weak base so you know it will be basic and you might guess pH 10 or so.
c) Mn^2+ + 2OH^- ==> Mn(OH)2(s)
d) Adding HNO3 would react with the OH^- of Mn(OH)2 to product H2O and the ppt would dissolve because the OH^- has been decreased.
1. a) What species are present in a 3.0 M NH3 solution? b) Qualitatively, what is the pH of this solution? c) What reaction(s) might occur when this solution is added to a solution containing Mn2+? d) What effect will adding HNO3 to the solution from part c have? Write any chemical equations that explain your answer.
3 answers
is there any way to calculate B?
Yes if you have Kb.
The problem says "qualitatively" which means make the best educated guess. Qualitative in that sense always means to make the best estimate.
If you know Kb (which is 1.8E-5) you can calculate it as follows:
..........NH3 + HOH ==> NH4^+ + OH^-
I.........3..............0........0
C........-x..............x........x....
E.........3-x............x........x
Kb = (NH4^+)(OH^-)/(NH3)
1.8E-5 = (x)(x)/(3-x). To get rid of a quadratic equation we usually assume 3-x = 3 (and in this case it is essentially true and making the assumption it will not change the answer) then
5.4E-5 = x^2
x = 0.0073 = (OH^-)
So pOH = -log(0.0073) = 2.13 and
pH + pOH = pKw = 14 so
pH = 14-2.13 = about 11.9
The problem says "qualitatively" which means make the best educated guess. Qualitative in that sense always means to make the best estimate.
If you know Kb (which is 1.8E-5) you can calculate it as follows:
..........NH3 + HOH ==> NH4^+ + OH^-
I.........3..............0........0
C........-x..............x........x....
E.........3-x............x........x
Kb = (NH4^+)(OH^-)/(NH3)
1.8E-5 = (x)(x)/(3-x). To get rid of a quadratic equation we usually assume 3-x = 3 (and in this case it is essentially true and making the assumption it will not change the answer) then
5.4E-5 = x^2
x = 0.0073 = (OH^-)
So pOH = -log(0.0073) = 2.13 and
pH + pOH = pKw = 14 so
pH = 14-2.13 = about 11.9
0 answers