Question

How much does the pH of 20.0 mL of a 0.75 M ammonium chloride (NH4Cl) solution change upon the addition of 15.0 mL of 1.0 M NaOH? Ka for ammonia (NH4+) 5.69 x 10-10?

Answer 1

Calculate initial pH:

NH4+ dissociates as:

NH4+ -----> H+ + NH3

0.75 0 0

0.75-x x x

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

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.69*10^-10)*0.75) = 2.066*10^-5

since c is much greater than x, our assumption is correct

so, x = 2.066*10^-5 M

So, [H+] = x = 2.066*10^-5 M

use:

pH = -log [H+]

= -log (2.066*10^-5)

= 4.68

Calculate pH after adding NaOH:

Given:

M(NH4+) = 0.75 M

V(NH4+) = 20 mL

M(NaOH) = 1 M

V(NaOH) = 15 mL

mol(NH4+) = M(NH4+) * V(NH4+)

mol(NH4+) = 0.75 M * 20 mL = 15 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 1 M * 15 mL = 15 mmol

We have:

mol(NH4+) = 15 mmol

mol(NaOH) = 15 mmol

15 mmol of both will react to form NH3 and H2O

NH3 here is strong base

NH3 formed = 15 mmol

Volume of Solution = 20 + 15 = 35 mL

Kb of NH3 = Kw/Ka = 1*10^-14/5.69*10^-10 = 1.757*10^-5

concentration ofNH3,c = 15 mmol/35 mL = 0.4286M

NH3 dissociates as

NH3 + H2O -----> NH4+ + OH-

0.4286 0 0

0.4286-x x x

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

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.757*10^-5)*0.4286) = 2.744*10^-3

since c is much greater than x, our assumption is correct

so, x = 2.744*10^-3 M

[OH-] = x = 2.744*10^-3 M

use:

pOH = -log [OH-]

= -log (2.744*10^-3)

= 2.5615

use:

PH = 14 - pOH

= 14 - 2.5615

= 11.44

So,

pH change = 11.44 - 4.68

= 6.76

Answer: 6.76