Monday

December 22, 2014

December 22, 2014

Posted by **Kelsy** on Tuesday, April 8, 2008 at 1:45pm.

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 -
**DrBob222**, Tuesday, April 8, 2008 at 3:08pm**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 -
**Kelsy**, Tuesday, April 8, 2008 at 3:09pmNo Sir, thank you :)

**Answer this Question**

**Related Questions**

Maths - There was a question: A gallon jug of milk is 3/4 full. After breakfast...

math - how can you get the answer 16 using these numbers 4,3,2,1=16 you can not ...

Math Algebra - Question:: You are about to take a test that contains questions ...

chemistry - A 1.325 g sample of an unknown vapor occupies 386 mL at 114 celsius...

Chemistry(Dr.Bob222) - What does it mean for something do to be neutralized? Is ...

Microbiology- Biology 205 - Can someone please help me with this question.... ...

math - just an algebra question. how would you solve: (3x^2 +5)(3x^2 +5) ^ means...

URGENT Spanish help please. - ok in the example it shows: ¿Comprendes espanol? ...

(19) Chemistry - Science (Dr. Bob222) - Use the following information to answer ...

Chemistry - Hello Jiskha, I'm hoping you are able to help me with my homework. ...