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.
Achieve Week 2 Q5
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Re: Achieve Week 2 Q5
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!
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Re: Achieve Week 2 Q5
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.
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.
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- Posts: 37
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Re: Achieve Week 2 Q5
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!
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!
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