Question

calculate the pH and the concentrations of all species in a saturated solution (16mg/100mL) of C21H22N2O2 which is a weak base (kb=1.8X10^-6)

Answer 1

MW C21H22N2O2 = 334.419 g/mol

mol in 16 mg --> mass/MW = (16*10^-3)/334.419 = 0.000047844 mol

[C21H22N2O2 ] = mol/V = (0.000047844) / (0.1) = 0.00047844

let the base be:

C21H22N2O2

there are free OH- ions so, expect a basic pH

B + H2O <-> HB+ + OH-

The equilibrium Kb:

Kb = [HB+][OH-]/[B]

initially:

[HB+] = 0

[OH-] = 0

[B] = M

the change

[HB+] = x

[OH-] = x

[B] = - x

in equilibrium

[HB+] = 0 + x

[OH-] = 0 + x

[B] = M - x

Now substitute in Kb

Kb = [HB+][OH-]/[B]

Kb = x*x/(M-x)

x^2 + Kbx - M*Kb = 0

x^2 + (1.8*10^-6)x - (0.00047844)(1.8*10^-6) = 0

solve for x

x = 2.85*10^-5

substitute:

[OH-] = 0 + x = 2.85*10^-5 M

pH = 14 + pOH = 14 + log(2.85*10^-5) = 9.45

pH = 9.45