Kw = KaKb. With this equation you can convert any Ka to Kb and the reverse.
Codeine (Cod), a powerful and addictive painkiller, is a weak base.
a) Write a reaction to show its basic nature in water. Represent the codeine molecule with Cod.
b) The Ka for its conjugate acid is 1.2 x 10^-8. What is Kb for the reaction written in a?
c) What is the pH of a .0020 M solution of codeine?
For a, I have:
Cod + H2O <==> HCod + OH-
For b, I'm a little confused. I know the conjugate acid is HCod. However, I don't know how to use that value to help me solve Kb.
I'm pretty sure I know how to solve C.
Any help is appreciated :)
how do u solve for C and b is basically the Kw(1X10^-14)/the Ka
Cod+ H20 <==> HCod+ OH-
0.002M x M x M
0.002-x +x +x
Kb=(HCod)(OH)/(Cod)=x^2/(0.002-x)
8.33x 10^-7=x^2/(0.002-x)
x=4.08x 10^-5 (theres [HCod])
pOH= PKb + log (HB/B-)
pOH= -log(8.33x 10^-7) + log(4.08x10^-5/0.002) = 4.38
POH + pH = 14 : 14- pOH = pH
14-4.38=
9.62
a) The reaction to show the basic nature of codeine in water can be represented as:
Cod + H2O ⇌ HCod + OH-
b) To find Kb for the reaction, we can use the relationship Kw = Ka * Kb. Rearranging the equation, we have:
Kb = Kw / Ka
Given that the Ka for HCod is 1.2 x 10^-8, we need to determine the value of Kw.
The value of Kw is the equilibrium constant for the self-ionization of water, which is equal to 1.0 x 10^-14 at 25°C.
Now, substituting the values into the equation:
Kb = (1.0 x 10^-14) / (1.2 x 10^-8)
Kb ≈ 8.3 x 10^-7
c) To find the pH of a 0.0020 M solution of codeine, first, we need to determine the concentration of hydroxide ions (OH-) using Kb and the initial concentration of codeine.
From the reaction in part a, we know that the codeine concentration is equal to the concentration of the conjugate acid, [HCod] = [Cod]. Thus, [Cod] = 0.0020 M.
Using the Kb value calculated in part b, we can set up an ICE table to determine the concentrations of HCod and OH- at equilibrium:
Cod + H2O ⇌ HCod + OH-
Initial: 0.0020 0 0
Change: -x +x +x
Equilibrium:0.0020-x x x
Since the stoichiometric coefficient of OH- is 1, the concentration of OH- at equilibrium is also x.
To solve for x, we can use the Kb expression:
Kb = [HCod][OH-] / [Cod]
Substituting the values:
8.3 x 10^-7 = x * x / (0.0020 - x)
Assuming x is much smaller than 0.0020, we can simplify the equation:
8.3 x 10^-7 ≈ x * x / (0.0020)
Solving for x², we get:
x² ≈ 8.3 x 10^-7 * 0.0020
Taking the square root of both sides:
x ≈ √[(8.3 x 10^-7) * 0.0020]
x ≈ 1.82 x 10^-4
Now, we can calculate the pOH:
pOH = -log10[OH-] = -log10(1.82 x 10^-4) ≈ 3.74
Finally, we can find the pH using the relation:
pH = 14 - pOH = 14 - 3.74 ≈ 10.26
So, the pH of a 0.0020 M solution of codeine is approximately 10.26.