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

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.

No Sir, thank you :)

Based on the information provided, it seems like you are trying to calculate the pH of a 0.001 mol/L solution of strychnine (SN) given its Kb value of 1.0 x 10^-6.

To solve this problem, you need to set up an equilibrium expression using the Kb expression:

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

Let's assume that the concentration of SNH+ is represented by x, the concentration of OH- is also represented by x, and the concentration of SN is 0.001 - x (since the total concentration is 0.001 mol/L).

Now, substitute these values into the Kb expression:

1.0 x 10^-6 = (x)(x)/(0.001 - x)

Solving this equation will give you the value of x, which represents the concentration of OH-:

x = 0.0000316226

To find the pH, you need to take the negative logarithm (base 10) of the concentration of OH-:

pOH = -log(0.0000316226) = 4.5

Since pH + pOH = 14 (at 25°C), you can calculate the pH:

pH = 14 - 4.5 = 9.5

So, the pH of a 0.001 mol/L solution of strychnine is approximately 9.5.

However, it seems like there might be some mistakes in the calculations you provided.

In the first calculation, your result of 4.5 for pH appears to be correct, following the correct steps.

In the second calculation, you used the quadratic formula to find the concentration of OH-. However, there are some mistakes in your calculation. The correct discriminant would be sqrt((-0.001)^2 - 4(1)(0.000001)), which is positive. Therefore, you should have two real solutions for x. However, it looks like you calculated the solutions incorrectly.

Overall, I would recommend following the correct steps and double-checking your calculations to ensure accuracy.