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

Answer:

pH = 8.77

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:

$pH = pKa + log\frac{[NH3]}{[NH4Cl]}$

$pH = 9.25 + log\frac{[0.03]}{[0.09]}=8.77$