Question

Determine the pH, pOH, and the concentration of all species at equilibrium for each of the following solutions.

c) A solution containing 0.042 M ammonia (pKb = 4.74)

d) A solution containing 0.024 M ammonium chloride e) The pH of a 0.125M solution of HCN is 2.87. Calculate Ka for HCN.

f)A weak monoprotic acid has a Ka of 1.6 X 10-5. Determine the pH, pOH, and the concentration of all species at equilibrium if the initial concentration of acid is 0.213.

Answer 1

c)

Ammonia, NH3, solution of 0.042 M

NH3 + H2O = NH4+ + OH-

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

Kb = 10^-pKb = 10^-4.74 = 1.8*10^-5

initially

[NH4+] = 0

[OH-] = 0

[NH3] = 0.042

in equilibrium,

[NH4+] = 0 + x

[OH-] = 0 + x

[NH3] = 0.042-x

substitute in Kb

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

1.8*10^-5 = x*x/(0.042-x)

x = OH- = 8.6*10^-4

pOH = -log(8.6*10^-4) = 3.0655

pH = 14-pOH = 14-3.065

pH = 10.935

d)

for ammonium chloride, NH4Cl = NH4+ + Cl-

then

NH4+ + H2O = H3O+ + NH3

Ka = [H3O+][NH3]/[NH4+]

get Ka via:

pKa + pKb = 14

pKa = 14-`Kb = 14-4.75 = 9.25

Ka = 10^-pKa = 10^-9.25 = 5.55*10^-10

initially

[H3O+] = 0

[NH3] = 0

[NH4+]= 0.024

in equilibrium

[H3O+] = x

[NH3] = x

[NH4+]= 0.024+ x

substitute in Ka

5.55*10^-10 = x*x/(0.024-x)

x = H3O+ = 3.65*10^-6

pH = -log(H) = -log(3.65*10^-6) = 5.437

e)

calcualte Ka for HCN

HCN = H+ CN-

Ka = [H+][CN-]÷[CN-]

[H+] = x

[CN-] = x

[CN-] = 0.125 - x

calculate H+ from pH

[H+] = 10^-pH = 10^-2.87 = 0.00134896 M

then, x = 0.00134896

so

[CN-] = x = 0.00134896

[CN-] = 0.125 - 0.00134896 = 0.12365

substitute in Ka

Ka = (0.00134896*0.00134896)/(0.12365) = 0.00001471648

Ka = 1.47*10^-5

f)

Ka = 1.6*10^-5 if HA = 0.214

HA = H+ + A-

Ka = [H+][A-]/[HA]

1.6*10^-5 = x*x/(0.213-x)

x = 0.001838

[HA] = M-x = 0.213-0.001838 = 0.211162 M

[A-] = x = 0.001838

[H+] = 0.001838

pH = -log(H+)I = -log(0.001838) = 2.735

pOH = 14-pH = 14-2.735 = 11.265

[OH-] = 10^-pOH = 10^-11.265 = 5.4325*10^-12