reaction of Zn(ll), E^percentage = -0.763 V; reduction of Cu(ll), E^percentage =

Question

reaction of Zn(ll), E^percentage = -0.763 V; reduction of Cu(ll), E^percentage = 0.34 V Kb for acetate = 5.6 x 10^-10 gQiuOMy product for silver bromide = 5.0 x 10^-13 iQbase SI energy units, the universal gas constant = 8.314 JK^-1 mol^-1 batt-life of C-14 = 5715y; Initial average activity of C-14 = 14.9 dpm g^-1 WaVQlfrngtte of colored light, in nm (assume 2 sig figs): red = 700; orange = 600; yellow = 560; green = 510; blue = 450; violet = 420 Planck's constant = 6.63 x 10^-34 J.s per photon Faraday constant = 96485 C/rnol In the titration of 50.0 mL of 0.15 M acetic acid (K, = 1.8 x 105) with 0.10 M barium hydroxide, calculate the pH After addition of 50.0 mL of the base After addition of 100.0 mL of base Calculate the solubility of silver bromide (K 5.0 x 10^-13 n) in a solution of 0.010 M barium bromide

Explanation / Answer

The titration is between a weak acid acetic acid and a strong diacidic base barium hydroxide.

The reaction is

HA + Ba(OH)2  --------> BaA2  + H2O

Q1) Before the tiratration begins

The solution has only acetic acid , whose pH is given by

pH = 1/2 pKa - 1/2 log C

= 1/2(4.74) -1/2 log 0.15

= 2.7889

Q2) After addition of 10mL of base

   HA + Ba(OH)2  --------> BaA2  + H2O

50x0.15 10x 0.1 0 0 initial mmoles

= 7.5 = 1

7.5 2 0 0 initial meq

5.5 0 2 2 after reaction meq

5.5 0 1 1 mmoles after reaction

Now the solution behaves as a buffer as it has both conjugate base and weak acid in the solution

pH of the buffer is given by Henderen equation as

pH = pKa + log [conjugate base]/[acid]

= 4.74 + log [0.2]/[5.5] = 3.300

Q3) after addition of 50mL of base

  HA + Ba(OH)2  --------> BaA2  + H2O

50x0.15 50x 0.1 0 0 initial mmoles

7.5 10 0 0 initial meq

  0 2.5 7.5 7.5 after reaction meq

Since the solution contains a strong base and a salt the pH is decided by the strong base only.

Thus [OH-] in slution = 2.5/ 100=0.025 M

pOH = -log [OH-] = -log 0.025= 1.6020

and thus pH = 14-1.6020 = 12.3980

Q4) AFter adding 75 mL of base

  

   HA + Ba(OH)2  --------> BaA2  + H2O

50x0.15 75x 0.1 0 0 initial mmoles

7.5 15 0 0 initial meq

  0 7.5 7.5 7.5 after reaction meq

Thus [OH-] in slution = 7.5/ 125=0.06 M

pOH = -log [OH-] = -log 0.06= 1.2218

and thus pH = 14-1.2218 = 12.778

Q5) after addition of 100mL base

HA + Ba(OH)2  --------> BaA2  + H2O

50x0.15 100x 0.1 0 0 initial mmoles

7.5 20 0 0 initial meq

  0 12.5 7.5 7.5 after reaction meq

Thus [OH-] in slution = 12.5/ 150=0.0833 M

pOH = -log [OH-] = -log 0.0833= 1.079

and thus pH = 14-1.079 = 12.920

Q6)

For silver bromide Ag Br(s)<-----> Ag+ + Br-

  Ksp = 5.0 x 10-13 = [Ag+] [Br-]

In 0.01 M barium hydroxide solution [OH-] = 2x 0.01 = 0.02 M

now

Ag Br(s)<-----> Ag+ + Br-

- x 0.02 equilibrium concentrations

Thus Ksp = 5.0x 10-13 = x . 0.02

Thus x= 2.5x 10-12M

Thus the solubility of AgBr in 0.01M barium hydroxide is 2.5x 10-12M

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