Sapling Acids/base equilibrium #5

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Melody Haratian 2J
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Sapling Acids/base equilibrium #5

Postby Melody Haratian 2J » Sun Jan 24, 2021 10:42 am

Hi guys,
I’m having some trouble going through #5 on the acids/base sapling homework. The problem is:
The Kb for an amine is 1.118 * 10-5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.757? Assume that all OH- came from the reaction of B with H2O.

Alison Perkins 2B
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Re: Sapling Acids/base equilibrium #5

Postby Alison Perkins 2B » Sun Jan 24, 2021 10:47 am

From the pH, you can solve for pOH because pH + pOH = 14.
From there, you can make an ICE table, and solve for your x with the Kb equation.

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Re: Sapling Acids/base equilibrium #5

Postby EmilyGillen_1A » Sun Jan 24, 2021 11:43 am

In the problem you're given the pH, but you're also given Kb and asked to evaluate a basic reaction, so the first thing you need to do it find the pOH by using: pH + pOH = 14.

Once you have the pOH, you can use the knowledge that pOH = -log (OH- concentration). Using this, you get a 10^x value (depending on what unique information your problem gave you), and that value represents the concentration of OH- produced by the reaction.

Now that you have [OH-], you need to find the original amount of the base, so you use the given Kb value and the amount of OH- (that you just found) to use P/R = K and find original amount of base present. ([BH+][OH-]/[B] = Kb)

Once you find the concentration of base reactant, you put the [OH-]/ ([OH-][B]) which will give you the percent ionization. (You have to add the 2 together on the denominator because they both contribute to the concentration, and if you didn't add in the [B] you would just get 1, which isn't true.)

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