If you know Kb for ammonia, NH3, you can calculate the equilibrium constant, Ka,
ID: 869294 • Letter: I
Question
If you know Kb for ammonia, NH3, you can calculate the equilibrium constant, Ka, for the following reaction: NH4 NH3 + H+ by the equation: Ka = Kw/Kb Ka = Kw/Kb Ka = Kb/Kw Ka = 1/Kb none of these Points Earned: 1.0/1.0 the dihydrogenphosphate ion, H2P04-, has both a conjugate acid and a conjugate base. These are, respectively: HPO42-, PO43- HPO42-, H3PO4 H3PO4, PO43- H2PO4-, HP042- H3PO4, HPO42 Points Earned: 1.0/1.0 A0.33-mol sample of a diprotic acid, H2A, is dissolved in 250ml of water the K01 of this water is 10,10-5 and K02 is 1.0 10-10calculate the concentration of the A-2 in this solution.Explanation / Answer
1) NH3 is a base so we can write base dissociation constant Kb
NH3 + H+ ----------------------> NH4+
Kb = [ NH4+]/[NH3][H+]
after accepting proton NH3 base converts NH4+ acid.
now NH4+ dissociation constant caln be written
NH4+ ---------------------------> NH3 + H+
Ka = [NH3][H+]/[NH3]
if Ka given we can calculate kb and if kb given we can calculate Ka by using the relation
Kw = Ka x Kb
2)
H2PO4- ------------------------> HPO4-2 + H+
( acid ) (conjugate base)
H2PO4- + H+ -------------------> H3PO4
( base ) (conjugate acid)
an acid has its conjugate base and base has its conjugate acid
acid -----------------------> conjugate base + H+
base + H+ -----------------------> conjugate acid
removal of proton from acid gives conjugate base .
addition of proton to base gives conjugate acid
3) H2A concentration = 0.33 x1000/250= 1.32M
H2A --------------------------> HA- + A-
1.32 0 0
1.32 -x x x
Ka= [ HA-][A-]/[H2A]
Ka1 = x^2/ (1.32-x)
1.0 x10^-5 = x^2/ (1.32-x)
x= 0.0036
[HA-] = 0.0036M
HA- -------------------> H+ + A-2
0.0036 0 0
0.0036-x x x
Ka2 = [ H+][ A-2]/[HA-]
1.0 x10^-10 = x^2/ (0.0036-x )
x= 1.0 x 10^-10
[ A-2] = x = 1.0 x 10^-10 M
second ionisation constant Ka2 always = [ A-2]
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