Achieve Week 2 Q5

[ACabello 1G]

I'm having trouble with the question because I don't know where to begin. Can someone explain where to at least begin?

The Kb for an amine is 3.963×10−5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.462? Assume that all OH− came from the reaction of B with H2O.

[kate 2h]

Because you are given the pH you can use the formula that relates pH and pOH (pOH=14-pH), and then take the antilog of the pOH and plug it back into the equation for K using the given K value and solve for B, hope that helps!

[RalphGuevarra]

Hi,

Since they give Kb and pH, we are going to use the given base to get pOH
pOH = 14 - pH (9.462) --> pOH = 4.538

then use that pOH to find the [OH-]=[BH+]
[OH-] = 10^(-4.538) = 2.897x10^-5

then use this equation to find the [B]:
Kb = ([OH-][BH+])/[B]

Now, you find [B], to determine the percentage of the amine divide the [BH+] by formal [B] (which is the sum of [B] and [BH+], then multiply by 100.

[ElizabethTopalian 2K]

Would we not need initial molarity also to solve this problem?

[Mari Brielle Mailed 1B]

In response to Elizabeth's follow-up question:

We'd have to solve for the initial molarity using the relationship Kb = [HB+][OH-]/[B]. Since there is a 1:1 ratio between [OH-] and [BH+], we can say that [OH-] = [BH+]; the same value will be squared in the numerator of the equation above. This concentration can be found by finding the pOH and taking the inverse log. In this case, [OH-] = [BH+] = 10^(-4.538). Then you just solve for [B], which is the initial molarity.

After that, you can just follow the other steps described in Ralph's post to reach your final answer!