Question

Formic Acid is secreted by ants (HCO₂H). Calculate the pH of a 0.0025M solutions of formic acid. (Ka = 3,5 x 10^-8)

b) calculate the pH of a 15.0 M aqueous solution of NH₃ . (K_b = 1.8 x 10^-5)

c) the K_sp value for lead(III)oidide, Pb₂, is K_sp = 1.4 x 10^-8 at 25 ^o C, calculates its solubility at 25^oC

Answer 1

a)

HCO2H dissociates as:

HCO2H -----> H+ + HCO2-

2.5*10^-3 0 0

2.5*10^-3-x x x

Ka = [H+][HCO2-]/[HCO2H]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((3.5*10^-8)*2.5*10^-3) = 9.354*10^-6

since c is much greater than x, our assumption is correct

so, x = 9.354*10^-6 M

So, [H+] = x = 9.354*10^-6 M

use:

pH = -log [H+]

= -log (9.354*10^-6)

= 5.029

Answer: 5.03

b)

NH3 dissociates as:

NH3 +H2O -----> NH4+ + OH-

15 0 0

15-x x x

Kb = [NH4+][OH-]/[NH3]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.8*10^-5)*15) = 1.643*10^-2

since c is much greater than x, our assumption is correct

so, x = 1.643*10^-2 M

So, [OH-] = x = 1.643*10^-2 M

use:

pOH = -log [OH-]

= -log (1.643*10^-2)

= 1.7843

use:

PH = 14 - pOH

= 14 - 1.7843

= 12.2157

Answer: 12.22

c)

At equilibrium:

PbI2 <----> Pb2+ + 2 I-

   s 2s

Ksp = [Pb2+][I-]^2

1.4*10^-8=(s)*(2s)^2

1.4*10^-8= 4(s)^3

s = 1.52*10^-3 M

Answer: 1.52*10^-3 M