Question

What is the ph of a solution containing 0.12mol/l of nh4cl and 0.03 mol/l of naoh (pka is 9.25)?

Answer 1

Given the following:

Concentration of residual (NH4)+ = 0.12 - 0.03 = 0.09 mol / L
Concentration of created NH3 = 0.03 mol / L

The (NH4) + reacts with all (OH)- to form NH3 and H2O ( Water )

The resultant solution is NH4+ / NH3 buffer system in which (NH4)+ is acid and the NH3 is base or as called the salt.

For a buffer system :

pH = pKa + Log (acid) / (salt)
= 9.25 + Log [ (NH4)+ ] / [ NH3 ]
= 9.25 + Log [ 0.09 ] / [ 0.03 ]
= 9.25 + Log 3
= 9.25 + 0.4771
pH = 9.7271 or 9.73 which is rounded up to 2 decimal places

Answer 2

pH = 8.77

Step-by-step explanation:

The reaction between Ammonium chloride and sodium hydroxide yield the following products depicted in the ICE table below:

`NH_(4)Cl + NaOH\rightarrow NH_(3) + H2O + NaCl`

Initial 0.12 M 0.03 M - - -

Change -0.03 M -0.03 M +0.03 M

Eq (0.12-0.03) (0.03-0.03) +0.03 M

Therefore,

`[NH_(4) Cl] = 0.12-0.03= 0.09 M\\`

`[NH_(3)] = 0.03 M`

This is a buffer with pKa = 9.25

Based on the Henderson-Hasselbalch equation:

`pH = pKa + log([NH3])/([NH4Cl])`

`pH = 9.25 + log([0.03])/([0.09])=8.77`