Buffer consists of ....

Question

A buffer consists of 0.26 M NH4Cl and 0.36 M NH3. What is the pH of the buffer after 0.006 moles of KOH are added to 0.20 L of this buffer solution? Kb for NH3 is 1.8E-5

Answer 1

We know that for buffer calculations, we can use

[H3O+] = Ka * (acid/base)

Remember that...

Kw = Ka*Kb = 1E-14, so

Ka = 1E-14/Kb = 1E-14/1.8E-5 = 5.56E-10

Then,

By adding a strong base, we can then create our ICE table using moles, so

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

0.006.. 0.052 ........--....... 0.072

-0.006 -0.006 ........-- .......+0.006

0 .........0.046 .......-- ........0.078

(note: assume all OH- is used up)

Thus,

[H3O+] = Ka * (acid/base)

[H3O+] = 5.56E-10 * (0.046/0.078)

[H3O+] = 3.27897E-10M

pH = -log[H3O+] = 9.48

[Answer: see above]