chemistry

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Hy Dr Bob222 this will be the last time I will bother you about this type of question. Can you please tell what I am doing wrong?
This is the best answer I could come up with:

Question: Strychnine, C21H22N2O2(aq) is a weak base but a powerful poison. Calculate the pH of a 0.001 mol/L solution of strychnine. The Kb of strychnine is 1.0 x 10^-6.

Calling strychnine SN,
SN + HOH --> SNH+ + OH-

Kb = (SNH+)(OH-)/(SN)
(SNH+) = x
(OH-) = x
(SN) = 0.001 - x

Plug Kb and solve for x = (OH-)

1.0 x 10^-6 = (x)(x)/0.001 - x
x^2 = 1.0 x 10^-9
x = 0.000031622
=-log (0.000031622)
=4.5

pH = 4.5

OR

x^2 + 0.001-x + 0.000001

Using quadratic formula gives

-0.001+0.0015 or -0.001-0.0015
= 0.0005 =negative answer

pH=-log^0.005
=2.3
pH = 2.30

  • chemistry -

    Thanks for showing your work. It helps us identify the problem(s).

    Calling strychnine SN,
    SN + HOH --> SNH+ + OH-

    Kb = (SNH+)(OH-)/(SN)
    (SNH+) = x
    (OH-) = x
    (SN) = 0.001 - x

    Note this next step. Note that x = (OH^-) and not (H^+).
    Plug Kb and solve for x = (OH-)

    1.0 x 10^-6 = (x)(x)/0.001 - x
    x^2 = 1.0 x 10^-9
    x = 0.000031622
    =-log (0.000031622)
    =4.5

    pH = 4.5
    Two things here.
    #1. If you let 0.001-x = 0.001, then x = 3.162 x 10^-5 and your answer is correct. That is the simplifying assumption and causes a small error. I don't know what your prof allows but this is a little over 3%; i.e., (3.162 x 10^-5/0.0010)*100 = 3.16%.

    #2. If s/he allows this, then this is the (OH^-) (note it isn't H^+ and this is your largest error) and (OH^-) = 3.16 x 10^-5, pOH = 4.5 and pH = 14-4.5 = 9.5 so we are at pH = 9.5 if your prof allows that small error.

    OR

    Next, on the quadratic formula, that isn't done correctly. I have shown that below in bold.
    x^2 + 0.001-x + 0.000001

    Using quadratic formula gives

    -0.001+0.0015 or -0.001-0.0015
    = 0.0005 =negative answer

    pH=-log^0.005
    =2.3
    pH = 2.30
    Starting with the original formula, we have (and I will use the E notation so as not to confuse x the unknown with x the times sign),
    [(x)(x)/(0.001-x)] = 1E-6
    x^2 = (0.001-x)*1E-6
    x^2 = 1E-9 - 1E-6x
    x^2 + 1E-6x - 1E-9 = 0
    Solving the quadratic formula we get, and I will leave that for you to do but I get x = 3.11 x 10^-5 M = (OH^-)
    pOH = 4.507 which I would round to 4.51 and pH = 9.49.

    You can see that the difference between 4.50 and 9.50 versus 4.51 and 9.51 is quite small. You know what the rules are in your class so it will be up to you to use the one that you should use but this gets the problem worked for you and I hope you can see why you had answers that didn't agree when worked by different methods. Please let me know if there is anything else. We are here to help. Again, thanks for showing your work. It makes the difference between guessing what you did wrong and knowing what you did wrong.

  • chemistry -

    No Sir, thank you :)

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