Achieve #5

[Kaitlin Joya 1I]

The Kb for an amine is 4.169 x 10^-5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.709? Assume that all OH- came from the reaction of B with H2O.

How do you use Kb to find [B]? I'm confused by what to do after finding [OH-] = [BH+]. If the equation for the ICE table should be B + H2O -> OH- + BH+ , how do you find [B] if x is being subtracted from it?

[Mia Glinn 1I]

You have to begin by using the pH to solve for the pOH (pH+pOH=14). With the pOH, we can solve for the concentration of OH-, which is one of the products in this chemical reaction. With the numbers in my #5, it would be:

14-9.481=pOH
pOH=4.519

[OH-]=10^-4.519
[OH-]=0.00003

Then you would solve the ICE table with X being the initial concentration of B. Finally, you can find the percentage protonated with that initial concentration and the concentration of BH+ (which is the same as OH-].

[Kaitlin Joya 1I]

You have to begin by using the pH to solve for the pOH (pH+pOH=14). With the pOH, we can solve for the concentration of OH-, which is one of the products in this chemical reaction. With the numbers in my #5, it would be:

14-9.481=pOH
pOH=4.519

[OH-]=10^-4.519
[OH-]=0.00003

Then you would solve the ICE table with X being the initial concentration of B. Finally, you can find the percentage protonated with that initial concentration and the concentration of BH+ (which is the same as OH-].

Do you mean the ICE table would be like this? I am confused by how the equilibrium concentration for B would be known to find [B]initial.

[alexjung1A]

You have to begin by using the pH to solve for the pOH (pH+pOH=14). With the pOH, we can solve for the concentration of OH-, which is one of the products in this chemical reaction. With the numbers in my #5, it would be:

14-9.481=pOH
pOH=4.519

[OH-]=10^-4.519
[OH-]=0.00003

Then you would solve the ICE table with X being the initial concentration of B. Finally, you can find the percentage protonated with that initial concentration and the concentration of BH+ (which is the same as OH-].

Do you mean the ICE table would be like this? I am confused by how the equilibrium concentration for B would be known to find [B]initial.

I think that the ICE table would be set up differently; the [B] we want to find is actually the FINAL concentration of [B] after the reaction, so there is no reason to use X at all in the ICE table. We have 0 initial concentration of OH- and BH+, so the 5.1 x 10^-5 should be the [OH-] and [BH+]. This way, we can set up the equation Kb = [OH-][BH+] / [B] then switch Kb and [B] so [B] = [OH-][BH+] / Kb. Now that we have the final [B] value, we can figure out how much of the B has been protonated to BH+ with the equation [BH+] / [BH+] + [B] * 100%. Basically, we divide amount of B protonated ([BH+]) by the total amount of B ([B] + [BH+]) in the final solution and multiply by 100 for the %.