Calculate the pH of a 0.10 M solution of aniline(C6H5NH2, Kb=3.8 x 10-10.)
A. 11.21 B.5.21 C. 9.42 D.2.79 E. 8.79
Answer : pH = 8.79 , (E) option is correct
explanation:
write down the equilibrium equation,
C6H5NH2 + H2O <======> C6H5NH3(+) + OH-
We have initial concentration of aniline = 0.10 M.
Constructing an ICE table.
C6H5NH2 | C6H5NH3(+) | OH- | |
---|---|---|---|
Initial | 0.10 | 0 | 0 |
Change | -x | +x | +x |
EQUILIBRIUM | 0.10-x | +x | +x |
Expression for Kb =[C6H5NH3+][OH-]/[C6H5NH2]
3.8*10^-10 = x*x/(0.10-x)
3.8*10^-10 = x^2 /0.10-x
x^2 + 3.8*10^-10x - 0.38*10^-10 =0
x = 6.16*10^-6 =[OH-]
pOH = -log [OH-] = -log [6.16*10^-6] = 5.21
pH = 14-pOH = 14 - 5.21
pH = 8.79
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