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
My work:
I called Strychnine SN
SN + HOH --> SNH+ + OH-
Kb = (SNH+)(OH-)/(SN)
(SNH+) = x
(OH-) =x
(SN) = 0.001-x
Plugging into Kb
Kb = (x)(x)/(0.001-x)
Can someone please solve for x = (OH-)?
Thank you
See above.
To solve for x (OH-), we need to rearrange the equation for Kb:
Kb = (x)(x) / (0.001 - x)
Multiplying both sides by (0.001 - x):
Kb(0.001 - x) = x^2
Expanding the left side:
0.001Kb - Kbx = x^2
Rearranging the equation:
x^2 + Kb x - 0.001Kb = 0
Now we have a quadratic equation in terms of x. Using the quadratic formula, we can solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = Kb, and c = -0.001Kb.
x = (-Kb ± √(Kb^2 - 4(1)(-0.001Kb))) / (2(1))
Simplifying further:
x = (-Kb ± √(Kb^2 + 0.004Kb^2)) / 2
x = (-Kb ± √(1.004Kb^2)) / 2
x = (-Kb ± Kb√1.004) / 2
Now, to determine whether to use the positive or negative sign, we need to consider the concentration of OH-. Since it is much smaller than the initial concentration of strychnine (0.001M), we can assume that x is much smaller than 0.001M. Therefore, x ≈ Kb√1.004.
Finally, we plug in the value of Kb (1.0 x 10^-6) into the equation:
x ≈ (1.0 x 10^-6)√1.004
x ≈ (1.0 x 10^-6) x 0.032
x ≈ 3.2 x 10^-8
So, the approximate concentration of OH- (x) in the 0.001M solution of strychnine is 3.2 x 10^-8 M.
To calculate the pH of the solution, we can use the equation: pH = -log10[OH-]
pH ≈ -log10(3.2 x 10^-8)
pH ≈ 7.5
Therefore, the approximate pH of the 0.001M solution of strychnine is 7.5.