Posted by Dee on Thursday, April 10, 2014 at 2:28pm.
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.
- Chemistry - DrBob222, Thursday, April 10, 2014 at 2:45pm
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^+
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.
- Chemistry - Dee, Thursday, April 10, 2014 at 2:51pm
Okay thank you so much!
- Chemistry - Dee, Thursday, April 10, 2014 at 3:03pm
Why do I subtract .9375 from H^+ instead of adding .9375 and then subtracting .9375 from NH4^+ instead of adding?
- Chemistry - DrBob222, Thursday, April 10, 2014 at 3:15pm
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.
- Chemistry - Derp, Thursday, April 10, 2014 at 5:50pm
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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