Write an equilibrium reaction for aniline dissolved in water. The pH for a 0.10
ID: 839887 • Letter: W
Question
Write an equilibrium reaction for aniline dissolved in water.
The pH for a 0.10 M solution of aniline is 9.46. Determine Kb for aniline. (You first have to determine [OH-])
Determine the pH of a solution that is 0.43 M B and 0.20 M BH+
Determine the molar solubility of each of the following, under the specified conditions.
AgI, in distilled water. Ksp= 8.5 X 10-17
Ag2CrO4 in a solution containing 1X 10-3 M CrO4-2
Fe(OH)3 in distilled water
Fe(OH)3 in water buffered at pH = 8.5
Determine the solubility of AgCl(s) in a solution that is 2.00 M in NH3 and 0.050 M in Cl-.
Explanation / Answer
1)
a) C6H5NH2 + H2O <--------> C6H5NH3+ + OH-
b) Reaction: C6H5NH2 + H2O <--------> C6H5NH3+ + OH-
Initial 0.1 0 0
Equilibrium 0.1-x x x
Kb = x2 / 0.1-x
we know, pH = 14 - pOH
x = [OH-] = 10-(14-9.46) = 10-4.56 = 3.64x10-4
Since, x << 0.1 , Kb = x2 / 0.1 = 1.325x10-8.
c) Reaction: B + H2O <--------> BH+ + OH-
Initial 0.43 0.2 0
Equilibrium 0.43-x 0.2+x x
Kb = x(0.2+x) / 0.43-x = 1.325x10-8
Since, x is very small , 0.2+x = 0.2 and 0.43-x =0.43
=> x(0.2) / 0.43 = 1.325x10-8
=> x = 2.848 x 10-8 = [OH-]
Hence, pOH = 14 - pH = -log(2.848 x 10-8) = 7.54
Therefore, pH = 14 - 7.54 = 6.46
2)
a) AgI + H2O <--------> AgOH + HI
Initial 1 0 0
equilibrium 1-x x x
Ksp = x2 / 1-x = 8.5 x 10-17
x = 9.22 x 10-9 mol / L is the solubility.
b) Ag2CrO4 + H2O -----------> 2AgOH + H2CrO4
Initial 0.001 0 0
equilibrium 0.001-x 2x x
Ksp = 2x2 / 0.001-x
8 x 10-12 = 2x2 / 0.001
=> x = 6.32 x 10-8 mol/L is the solubilty.
c) In distilled water, [OH-] = 10-7
For this, Ksp = [Fe3+][OH-]3 where [Fe3+] = x and [OH-] = 3x
=> 2 x 10-28 = x * 10-21
=> x = 2*10-7
d) pH = 8.5 => [H+] = 10-8.5 => [OH-] = 10-5.5 = 3.16 x 10-5
For this, Ksp = [Fe3+][OH-]3 where [Fe3+] = x and [OH-] = 3.16 x 10-5
=> 2 x 10-28 = x * (3.15*10-15)
=> x = 6.35*10-14
3) Adding both the equations,
AgCl + 2NH3 <--------> Ag(NH3)2+ + Cl-
Initial 1 0.2 0 0.05
equilibrium 1 0.2-2x x 0.05+x
K = Ksp*Kf = 2.56*10-3 = x(0.05+x) / 0.2-2x
Solving the quadratic eqation, x2 + 0.05x = 0.000512 - 0.00512x
=> x2 + 0.05512x = 0.000512
=> x = 8.11*10-3 mol/L
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.