Question

A 10.0 mL solution of 0.300 M NH3 is titrated with a 0.100 M HCl solution. Calculate the pH after the addition of 20.0 mL of the HCl solution.

(potentially useful info: Ka of NH4+ = 5.6 x 10−10)

Answer 1

a)

mmol of base initially = MV = 10*0.3 = 3 mmol of NH3

mmol of acid added = MV = 0.1*20 = 2 mmol of HCl

there is reaction so

NH3 + HCl = NH4+ + Cl-

note that

NH3 decrease, and NH4´incrases

mmol of NH3 left = 3-2 = 1 mmol of NH3 left

mmol of NH4+ produced = 0 + 2 = 2 mmol of NH4+

then, this is a buffer

A buffer is any type of substance that will resist pH change when H+ or OH- is added.

This is typically achieved with equilibrium equations. Both type of buffer will resist both type of additions.

When a weak acid and its conjugate base are added, they will form a buffer

The equations:

The Weak acid equilibrium:

HA(aq) <-> H+(aq) + A-(aq)

Weak acid = HA(aq)

Conjugate base = A-(aq)

Neutralization of H+ ions:

A-(aq) + H+(aq) <-> HA(aq); in this case, HA is formed, H+ is neutralized as well as A-, the conjugate

Neutralization of OH- ions:

HA(aq) + OH-(aq) <-> H2O(l) + A-(aq) ; in this case; A- is formed, OH- is neutralized as well as HA.

Now,

For the weak base equilibrium:

B(aq) + H2O(l) <-> BH+(aq) + OH-(aq)

Weak base = B;

Conjugate acid = BH+

Neutralization of OH- ions:

BH+(aq) + OH-(aq) <-> B(aq) + H2O(l); in this case, OH- is neutralized by BH+, as well as B is created

Neutralization of H+ ions:

B(aq) + H+(aq) <-> BH+(aq)

apply buffer equation

pH = pKa + log(NH3/NH4+)

substitute

pKa = -log(Ka) -log(5.6*10^-10) = 9..25

pH = pKa + log(NH3/NH4+)

pH = 9.25+ log(1/2)

pH = 8.948970