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pH of a mixed solution: is this right?


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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?

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  • 4 weeks later...

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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