Achieve #5

[905574994]

Hi guys,

Could someone help with this problem?
The Kb for an amine is 9.624×10−5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.680? Assume that all OH− came from the reaction of B with H2O.

I'm not really sure where to start.

Thanks!

[905744270]

Hey! There was actually already a really great discussion in Chem Community that helped me with this problem. Here's the link: viewtopic.php?f=49&t=89720

[Ally Mosher]

kb and ka are related. kb times ka is equal to 10^(-14). I hope this helps.

[Coline Luo 2E]

Hi! I think like most problems, you should start by writing out the equation for an amine reaction, which is essentially like a weak base. Since you are given pH, you can find [BH+] and [OH-]. Using these concentrations and Kb, you can find [B]. Now that you have [B] and [BH+], you can find the percentage protonated. Hope this helps!

[Kiana Rogers 1E]

First we can look at the equations:
B(aq)+H2O(l)↽−−⇀BH+(aq)+OH−(aq)
Kb=[BH+][OH−][B]

With being given the pH value, we can get the pOH value by doing: pOH = 14 - pH
We know that the pOH = -log[OH-], so we can find the [OH-] and [BH+] value, or the so called "X" value.

We then would make an ICE box to find the Binitial value.
To find the % protonated, we would do [BH+]/[Binitial] * 100%.

Hope this helps.

[Emily Wan 1l]

From the pH value, the pOH value can be calculated and consequently the equilibrium [OH-] concentration. From there, you can set up an ICE table to solve for the missing initial amine concentration. The percent protonated can then be calculated by dividing the equilibrium concentration over the initial concentration and multiplying by 100.

[Phoebe Ko 3E]

I believe a good place to start is to write out the chemical reaction equation. Since the question directly states "the reaction of B with H2O," we could write out a weak base equation of B + H2O <-> BH+ + OH-. From the given pH, we could calculate the pOH, as well as [OH-] and [BH+] (which would be equal to each other). Using an ICE chart and the given Kb value, we could then calculate the initial [B]. Lastly, with the initial [B] and [BH+] that we calculated earlier, we would be able find the percent protonated. I hope this was helpful!

[Madelyn_Rios_2c]

1. find pOH which is 14-pH.
2. find [BH+] which equals [OH-]:
[OH-] = 10^(-pOH)
3. find [B] by using Kb = ( [BH+][OH-] ) / [B]
4. to find the percentage of the amine that is protonated use:
( [BH+] / ( [BH+] + [B] )) x 100

[Saebean Yi 3E]

First, let's identify what the problem ultimately wants us to solve, which is percentage protonated. In this case, since we're dealing with bases, the protonated % should be [BH+] (at equilibrium) / [B] initial.
And again, since we're talking about bases, we want to talk about pOH. Since pH is given and we know 14 = pH + pOH, pOH can be easily derived.
Then write out the equation given. Should look like: B(aq)+H2O(l) <- -> BH+(aq)+OH−(aq). And then put this into the ICE table. But since we're already given the "final values" (pH), we are working bottom to up, rather than top to down.
pOH = -log [OH-] and from the ICE table, we know [OH-] = [BH+]. All that's left to do is finding [B] initial.
At this point, your ICE table should be mostly filled in except [B] initial. But since we know Kb, we can calculate it via Kb=[BH+][OH−]/[B]. Note that the [B] value is at equilibrium, so add the "change" from the ICE table to get [B] initial.

[Yalit Gonzalez 1A]

Step 1: Find pOH: 14-pH=pOH
Step 2: Find [OH-]: [OH-]=antilog^-pOH
Step 3: Find B: [OH-]*[BH+]/Kb. Note: [OH-]=[BH+]
Step 4: Find final answer: [BH+]/[BH+]+B *100