Question

A buffer is made by adding 0.12 mol NH4Cl and 0.15 mol NH3 to enough water to make 0.250 L of solution. (Ka NH4+/NH3= 5.6 x10−10)

a) Calculate the pH of the buffer solution

b) What is the pH after the addition of 10.00 mL of 0.1 M NaOH to 90.00 mL of this buffer solution? Compare to pH in part a.

c) What is the pH of a solution made by adding 10.00 mL of 0.1 M NaOH to 90.00 mL of pure water? Compare to pH in part b.

Answer 1

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,

pH = pKa + log(NH3/NH4+)

pKa = -log(Ka) = -log(5.6*10^-10) = 9.25

pH = 9.25 + log(NH3/NH4+)

a)

initially

pH = 9.25 + log(NH3/NH4+)

pH = 9.25 + log(0.15/0.12)

pH = 9.3469

b)

after

mmol of NaOH = MV = 0.1*10 = 1 = 10^-3

mol of NH3 = 90/250 * 0.15 = 0.054

mol of NH4+ = 90/250 * 0.12 = 0.0432

after OH- reacts

mol of NH3 = 0.054 + 10^-3 = 0.055

mol of NH4+ = 0.0432 -10^-3 = 0.0422

pH = 9.25 + log(NH3/NH4+)

pH = 9.25 + log(0.055/0.0422)

pH = 9.365

c)

after adding it to pure water

mol of NaOH = 10^-3

V total = 10+90 = 100 mL = 0.1

[NaOH] = mol/V = 10^-3 / (0.1) = 0.01

pOH = -log(0.01) = 2

pH = 14-2 = 12

pH = 9.365 vs 12

change in ph is drastic if no buffer is present