What is the final pH of a solution obtained by mixing 200 ml of 0.400 M NH3 with 300 ml of 0.100 M HCl? (Kb = 1.8 x 10^-5)

 

I used an ICE table and came up with 9.68. Is that correct?

 

Should I have used two ICE tables instead of one? The first in the one direction and the second going back in the other direction?

To begin with, there are initially (0.2 L)(0.4M) = 0.08 mol NH3. (0.3 L)(0.1M) = 0.03 mol HCL is added, which completely ionizes. The 0.3 mol H+ neutralizes 0.03 mol NH3. Thus (0.08 - 0.03 mol) NH3 = 0.05 mol of NH3.

 

The new concentration of NH3 is ( 0.05 mol ) / (0.2L + 0.3L) = 0.1M due to the volume of the HCL

 

The reaction is now NH3(aq) -> NH4+ + OH-

 

Kb = [NH4+][OH-]/[NH3] which yields 1.8 x 10^-5 = x^2 / (0.1-x)

 

Solving for x=0.00133 mol/L and x=[OH-]

 

-log [OH-] = pOH --> -log [0.00133] = 2.875

 

pOH + pH = 14 thus 14-2.875 = pH = 11.12

 

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