Achieve #5

[205819952]

The Kb for an amine is 3.297×10−5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.710 ? Assume that all OH− came from the reaction of B with H2O.
How do I solve this?

[305597516]

Hi! I had problems with this question as well, but was able to solve it. So essentially, you are looking for the initial molarity of the base, and to find it, you will need to create an ICE table. The chemical reaction should be written as:
B + H2O --> OH- + B+
The first thing to do is to find the concentration of OH-. You can do this by taking the given pH and subtracting it from 14 to find the pOH. You can then reverse log that answer to find the concentration. Because there is only 1 mole of B being turned to 1 mole each of products, both OH- and B+ will be the same concentration, so whatever you get for OH- is the same for B+.
Next, construct your ICE table. You can use the variable B to show the unknown initial value, and since we are assuming that the forward reaction is what the problem is asking, the initial values for OH- and B+ should be 0. For the change, you know that B will be decreasing to form the products, so it will decrease by x and the products will both increase by x. You already know what x is from finding the concentration of OH-.
You can then use the ice table to find B based on already knowing what x and Kb are; you can then use B and the protonated B+ product to find the percentage of the amine has been protonated. Hope this helped!

[Kai Tokiyeda 2K]

1. Find the pOH from the pH which is given to you. (14-9.710)
2. Use pOH value to find [OH-]. Use formula pOH = -log[OH]. You'll need to do inverse log, or basically 10^-pOH
3. Substitute your [OH-] value in the K_b formula. 3.297×10−5 = [OH-]²/[B]
4. Once you solve for [B] subtract [OH] from your answer. This will give you the initial concentration.
5. Solve for % protonated. Formula is [OH]/initial concentration x 100%

[kayleec1004]

First, find the pOH from the pH. Second, calculate 10-pOH. Third, use the Kb formula and replace the [OH-]. Fourth, solve for [B] and subtract [OH] from your answer to find the initial concentration. Finally, solve for the percentage protonated.

[Sean1F]

1. Find the pOH from the pH which is given to you. (14-9.710)
2. Use pOH value to find [OH-]. Use formula pOH = -log[OH]. You'll need to do inverse log, or basically 10^-pOH
3. Substitute your [OH-] value in the K_b formula. 3.297×10−5 = [OH-]²/[B]
4. Once you solve for [B] subtract [OH] from your answer. This will give you the initial concentration.
5. Solve for % protonated. Formula is [OH]/initial concentration x 100%

Kai you deserve to win the lottery for breaking this down into steps. thanks brah