The degree to which a weak base dissociates is given by the base-ionization cons

Question

The degree to which a weak base dissociates is given by the base-ionization constant, Kb. For the generic weak base, B

B(aq)+H2O(l)BH+(aq)+OH(aq)

this constant is given by

Kb=[BH+][OH]/[B]

Strong bases will have a higher Kb value. Similarly, strong bases will have a higher percent ionization value.

Percent ionization=[OH] equilibrium/[B] initial×100%

Strong bases, for which Kb is very large, ionize completely (100%). For weak bases, the percent ionization changes with concentration. The more dilute the solution, the greater the percent ionization.

Ammonia, NH3, is a weak base with a Kb value of 1.8×105

Part A

What is the pH of a 0.470 M ammonia solution?

Express your answer numerically to two decimal places.

Part B

What is the percent ionization of ammonia at this concentration?

Express your answer with the appropriate units.

Explanation / Answer

a)

This is a base in water so, let the base be NH3= "B" and NH4+ = HB+ the protonated base "HB+"

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^-5)x - (0.47)(1.8*10^-5) = 0

solve for x

x = 0.002899

substitute:

[HB+] = 0 + x = 0.002899 M

[OH-] = 0 + x = 0.002899 M

[B] = M - x = 0.47-0.002899= 0.467M

pH = 14 + pOH = 14 + log(0.002899 ) = 11.46

B)

% ionizatoin = BH+/B*100 %= 0.002899 /0.47 * 100 = 0.61680% = 0.62%

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