[ebonyscythe]

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(K_a) + 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

K_a for NH₄⁺ = 5.6 x 10^-10
K_b for NH₃ = 1.8 x 10^-5

[Borek]

C_diluted V_diluted = C_concentrated V_concentrated

Amount of substance doesn't change, volume does. Note that

V_diluted = V_concentrated + V_{added water}

[ebonyscythe]

So for solution 9 the concentration of NH₃ would be (.10)(10ml) = C_dil (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
          ----------------------------------
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?

[ebonyscythe]

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?

-----------

[HCl]=(.10M)(6.0mL)/16mL = .0375

[Borek]

So for solution 9 the concentration of NH₃ would be (.10)(10ml) = C_dil (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).

[ebonyscythe]

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!