quick CHEMISTRY question....

[abeal2]

Calculate the pH of ephedrine sulfate where the concentration of weak acid (ephedrine cation) is 10^-2 M. Kb for ephedrine base is 10^-5.

 

[James3shin]

fine Ka by using the relationship Ka(Kb)=Kw then use henderson-hasselbach

 

[killface]

We won't help you make meth.

 

[abeal2]

Originally posted by: James3shin
fine Ka by using the relationship Ka(Kb)=Kw then use henderson-hasselbach

so

Ka x Kb = Kw
Ka x 10^-5 = 10^-14
Ka = 10^-9
pKa = 9

pH = pKa + [A]/[AH]

i know pKa and [AH] (AH being 10^-2). how do i find [A]?

 

[drinkmorejava]

lol reminds me that i have a chem test I should study for tomorrow,

I'd just solve for pOH then subtract it from 14 to get pH.

 

[abeal2]

Originally posted by: drinkmorejava
lol reminds me that i have a chem test I should study for tomorrow,

I'd just solve for pOH then subtract it from 14 to get pH.

hehe, and how would i do that?

 

[da loser]

Originally posted by: abeal2

Originally posted by: James3shin
fine Ka by using the relationship Ka(Kb)=Kw then use henderson-hasselbach

so

Ka x Kb = Kw
Ka x 10^-5 = 10^-14
Ka = 10^-9
pKa = 9

pH = pKa + [A]/[AH]

i know pKa and [AH] (AH being 10^-2). how do i find [A]?

i think that it's equilibrium, and since AH <-> A + H; A=H

i haven't done this in a long time, but i believe Ka or Kb is just a rate constant of the equilibrium, and since it's a weak acid the reaction does not go completely. so they give you AH, and that allows you to calculate how far the reaction goes using Kb

 

[BD2003]

Originally posted by: da loser

Originally posted by: abeal2

Originally posted by: James3shin
fine Ka by using the relationship Ka(Kb)=Kw then use henderson-hasselbach

so

Ka x Kb = Kw
Ka x 10^-5 = 10^-14
Ka = 10^-9
pKa = 9

pH = pKa + [A]/[AH]

i know pKa and [AH] (AH being 10^-2). how do i find [A]?

i think that it's equilibrium, and since AH <-> A + H; A=H

i haven't done this in a long time, but i believe Ka or Kb is just a rate constant of the equilibrium, and since it's a weak acid the reaction does not go completely. so they give you AH, and that allows you to calculate how far the reaction goes using Kb

From what I can tell, it would have to be, otherwise there isnt enough info to solve the problem.