Amphetamine (C9H13N) is a weak base with a pKb of 4.2.
Calculate the pH of a solution containing an amphetamine concentration of 215 mg/L.
I calculated the molarity as 1.58*10^-3 and the Ka as 1.58*10^-10. I did ICE and i got the pH as 6.3, but it was wrong. Do you know what I did wrong?
I think you treated amphetamine as a salt. Its a base. pKa of 4.2 makes Kb = 6.3 x 10^-5. If we represent amphetamine as RN, then
RN + HOH ==> RNH^+ + OH^-
Kb = (RNH^+)(OH^-)/(RN)
6.3 x 10^-5 = (x)(x)/1/59 x 10^-3
solve for x which is (OH^-), convert to pOH, then to pH. I get something like 10.5 but that's a quickie. Check it out.
dont make sense
To calculate the pH of a solution containing amphetamine, you need to consider that amphetamine is a weak base. In this case, you correctly calculated the molarity as 1.58*10^-3 M, which is equivalent to 1.58*10^-3 mol/L.
However, it appears you made a mistake in calculating the Ka value. The Ka value represents the acid dissociation constant, which is the equilibrium constant for the ionization of a weak acid. Since amphetamine is a weak base, we need to determine the Kb value, which is the equilibrium constant for the ionization of a weak base.
The Kb value can be calculated from the Kw value (the ionization constant of water) and the Ka value using the following equation:
Kw = Ka * Kb
Since Kw is a constant equal to 1.0*10^-14 at 25°C, we can rearrange the equation to solve for Kb:
Kb = Kw / Ka
Plugging in the value of Ka (1.58*10^-10), we get:
Kb = (1.0*10^-14) / (1.58*10^-10)
= 6.33*10^-5
Now that we have the Kb value, we can proceed with calculating the pOH and pH.
pOH = -log10[Kb]
= -log10[6.33*10^-5]
= 4.2
Since pOH + pH = 14 (at 25°C), we can calculate the pH as follows:
pH = 14 - pOH
= 14 - 4.2
= 9.8
Therefore, the pH of a solution containing an amphetamine concentration of 215 mg/L is approximately 9.8.