Strychnine, C21H22N2)2(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
DrBob222, I am planning on obtaining a pHD in about nine years:P Can you give me some advice on how to be that 1% of the population who has a pHD?
chemistry - DrBob222, Sunday, April 6, 2008 at 12:38am
Start now!
Let's call strychnine SN.
SN + HOH ==> SNH^+ + OH^-
Kb = (SNH^+)(OH^-)/(SN)
(SNH^+) = x
(OH^-) = x
(SN) = 0.001 - x
Plug into Kb and solve for x = (OH^-)
pOH = - log(OH^-)
and pH = 14 - pOH
Kb = (SNH^+)(OH^-)/(SN)
= (x)(x) / 0.001 - x
= x^2 / 0.001 - x
How is the next step supposed to look?
Kb = (SNH^+)(OH^-)/(SN)
= (x)(x) / 0.001 - x
= x^2 / 0.001 - x
You haven't included Kb. Look in the problem, I know you posted it, and this last equation becomes
(x)(x)/(.001-x) = Kb. After adding the Kb into the mix, this is a quadratic equation unless the 0.001-x can be simplified to 0.001.
(x)(x)/(.001-x) = Kb
(x^2)/(0.001 - x) = 1.0 x 10^-6
So, in the form of ax^2 +bx + c = 0
this is
x^2 + 0.001x + 1.0 x 10^-6?
See my later post to this question. But what you have written isn't correct.
To calculate the pH of a 0.001 mol/L solution of strychnine, we need to use the Kb value of strychnine, which is 1.0 x 10^-6.
Let's call the concentration of SNH^+ as x and the concentration of OH^- as x. The concentration of SN (strychnine) would be 0.001 - x, as we are starting with a 0.001 mol/L solution.
The expression for Kb is Kb = (SNH^+)(OH^-)/(SN). Substituting the concentrations, we get Kb = (x)(x)/(0.001 - x).
To solve for x (OH^-), we need to set up an equilibrium expression and use the Kb value.
1. Kb = (x)(x)/(0.001 - x)
2. Rearrange the equation by multiplying both sides by (0.001 - x): Kb(0.001 - x) = x^2
3. Expand the equation: 0.001Kb - Kbx = x^2
4. Rearrange the equation and set it equal to zero: x^2 + Kb*x - 0.001Kb = 0
At this point, you can solve this quadratic equation using the quadratic formula or by factoring. Since the equation has a coefficient of 1 for x^2, it can be factored as follows:
(x + m)(x + n) = 0
Where m and n are the factors that add up to Kb, and when multiplied, give -0.001Kb.
After finding the values of m and n, you can write two equations:
x + m = 0
x + n = 0
Solve these two equations for x to find the concentration of OH^-.
Once you have the concentration of OH^-, you can calculate the pOH using the equation pOH = -log[OH^-]. Finally, to find the pH, you can use the equation pH = 14 - pOH.
Remember to review the calculations and check for any mistakes along the way. Good luck with your chemistry studies and your goal of obtaining a PhD!