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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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