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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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mickey4 mickey4
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9 years ago
Given that
•Ka for HCN  is 4.9×10-10   and
•Kb for NH3  is 1.8×10-5,
calculate
•Kb for CN- and
•Ka for NH4+
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#1
9 years ago
To find the Kb of HCN, just use the formula
Ka= Kw/Kb

Kw is always 1E-14 (water), so you just do 1E-14/4.9 E -10 to get a Ka of 2.04 E -5. With two sig figs, that would be 2.0 E-5.

Ka for NH4+ you would use the same formula, just switch it up a little...
Kb= Kw/Kb
You'll get 5.55E-10, using two sig figs that would be 5.6 E-10.

2.0 E-5, 5.6 E-10
wrote...
#2
9 years ago
For any conjugate acid-base pair, Ka*Kb=Kw=1.0*10-14

(1)For HCN and CN-:

equation showing Ka: HCN + H2O---- H3O+ + CN-

Ka=[H3O+][CN-]/[HCN];

equation of Kb: CN- + H2O ----- HCN + OH-

Kb=[HCN][OH-]/[CN-]

so Ka*Kb=[H3O+][OH-]=Kw=1.0*10-14

So Kb for CN- = (Kw)/Ka=(1.0*10-14)/(4.9*10-10)=2.0*10-5

(2)similarly, for NH4+ (conjugate acid) and NH3 (conjugate base):

Ka=Kw/Kb=(1.0*10-14)/(1.8*10-5)=5.6*10-10
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