Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: ebonyscythe on May 15, 2007, 01:34:13 PM
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We did a lab in class a while ago involving weak acid and base equilibria where we had to make up some .10 M solutions, measure pH and the later on calculate the pH and compare.
Here's the set of calculations I'm stuck on:
Solution 8: 25ml 0.10 M NH3 + 25ml 0.10 M NH4NO3
pH obtained in Lab - 9.17
Solution 9: 10ml soln 8 + 6 ml H2O
pH obtained in Lab - 9.24
Solution 10: 10ml soln 8 + 5 ml H2O +1ml 0.10 M HCl
pH obtained in Lab - 8.82
Solution 11: 10ml soln 8 + 6 ml 0.10 M HCl
pH obtained in Lab - 2.13
Solution 8 is a buffer solution, I know that... I'm having trouble going back to dilutions and calculating molarity. I need the concentrations of the solutions so I can calculate the pH of each on with the Henderson-Hasselbalch equation:
pH = -log(Ka) + log([base]/[acid])
How can I go about getting the concentrations for the HH equation?? I just can't seem to get started; I may be thinking too hard about it.
Any help is appreciated! Thanks.
Kris
Ka for NH4+ = 5.6 x 10-10
Kb for NH3 = 1.8 x 10-5
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Cdiluted Vdiluted = Cconcentrated Vconcentrated
Amount of substance doesn't change, volume does. Note that
Vdiluted = Vconcentrated + Vadded water
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So for solution 9 the concentration of NH3 would be (.10)(10ml) = Cdil (16ml) or .063?
Just want to make sure I'm on the right track.
And then for soln 10 is the concentration for NH3 is the same right? And the concentration of HCl is technically H3O+?
So I make a table to determine the concentrations I enter into the equation:
NH3 + H3O+ -----> NH4+ + H2O
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start .063 .0063 .063
change -.0063 -.0063 +.0063
left .056 0 .069
Then I put those concentrations into the HH equation to get:
pH = -log(5.6 x 10-10) + log(.057/.069) = 9.17
Is that correct?
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I must be doing something wrong because in Solution 11 the NH3 of the buffer is supposed to be all used up (hence the huge pH change to 2.13) and I'm not getting that. I'm calculating the molarity of the HCl(H3O+) to be .038 M; that's not enough to use up the .063 M NH3.
Can someone tell me what I'm doing wrong?
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[HCl]=(.10M)(6.0mL)/16mL = .0375
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So for solution 9 the concentration of NH3 would be (.10)(10ml) = Cdil (16ml) or .063?
Close, but solution 8 already contains diluted ammonia (volume was 25+25). Thus not 0.10.
And then for soln 10 is the concentration for NH3 is the same right?
Yes.
And the concentration of HCl is technically H3O+?
Yes.
So I make a table to determine the concentrations I enter into the equation:
NH3 + H3O+ -----> NH4+ + H2O
----------------------------------
start .063 .0063 .063
change -.0063 -.0063 +.0063
left .056 0 .069
Then I put those concentrations into the HH equation to get:
pH = -log(5.6 x 10-10) + log(.057/.069) = 9.17
Is that correct?
Something like that, just concentrations are wrong (see above).
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Ah, I see. I'll work on getting the right concentrations then. Thanks for pointing that out!
edit: .05 M for NH3 for the first dilution? And then .031M for the second? That seems right because then that would solve my problem for soln 11.
Thanks again!