Question

A buffer solution is prepared by mixing 50.0 mL of 0.300 M NH3 with 50 mL of 0.300 NH4Cl. The pKb of NH3 is 4.74.
NH3 + H2O-> NH4+ +OH-

7.50 mL of 0.125 M HCl is added to the 100 mL of the buffer solution. Calculate the concentration of NH3 and NH4Cl for the buffer solution. Calculate the pH of the solution.

Answers

millimols NH3 = 50 mL x 0.3M = 15.0
(NH3) = mmols/mL = 15.0/100 = 0.15M initially.

millimols NH4Cl = 50 x 0.3 = 15.0
(NH4Cl) = 15.0/100 = 0.15M initially

millimols HCl = 7.50 x 0.125 = 0.9375
pKb NH3 = 4.74; pKa = 14-4.74 = 9.26

I prefer to work with this equilibrium in mmols.
...........NH3 + H^+ ==> NH4^+
I..........15....0........15
add............0.9375..............
C......-0.9375..-0.9375...+0.9375
E.........???....???.......????
Subtraqct I-C = E line and substitute those values into HH equation solve for pH.
Final (NH3), (NH4Cl) = mmols each/total mL.
Post your work if you get stuck.
Okay thank you so much!
Why do I subtract .9375 from H^+ instead of adding .9375 and then subtracting .9375 from NH4^+ instead of adding?
With regard to H^+. You added 0.9375 mols H^+ to the buffer. ALL of it will react with the 15.0 mmols NH3 (H^+ is the limiting reagent so there won't be any of it left). With regard to the NH4+ you should add and my ICE chart shows that; however, I said subtract in my instructions. I should have said, "Add the I line to the C line algebraically) to arrive at the E line. That might have cleared up the H^+ question too if I had stated my instructions correctly.
I have one question. When you say divide by the total mL, did you include the 7.5mL of HCl that you added?

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