For question 5 on the second Achieve assignment, why do we use formal concentration on the bottom and not the initial. Context: the hint for that question in the assignment says that the formal concentration is used, and the formal concentration is the initial concentration of [B] + [HB+].
This is the question:
The Kb for an amine is 3.048×10−5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.832 ? Assume that all OH− came from the reaction of B with H2O.
Achieve Week 2, Question 5
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Re: Achieve Week 2, Question 5
Hello!
In this specific problem, we are not given the initial concentration. Instead, we are given the pH, which can be used to find the [OH-]. However, once we find the concentration of [OH-], we must use the KB given to find the initial concentration of B. In the equation for KB, the concentration of B is represented as [B]-x where x is [OH-]. As a result, the denominator of the percentage must be the initial concentration which is found by adding the OH[sup]-[/sub] to the equilibrium concentration of B. I hope this helps!
In this specific problem, we are not given the initial concentration. Instead, we are given the pH, which can be used to find the [OH-]. However, once we find the concentration of [OH-], we must use the KB given to find the initial concentration of B. In the equation for KB, the concentration of B is represented as [B]-x where x is [OH-]. As a result, the denominator of the percentage must be the initial concentration which is found by adding the OH[sup]-[/sub] to the equilibrium concentration of B. I hope this helps!
Re: Achieve Week 2, Question 5
It's helpful to think of this problem as solving it backwards that usual. We can easily get the [OH-] from the pH. Then from there we can plug everything into the Kb equations and rather than unknown x, we have unknown B. Solve for B and then simply use that and [OH-] to answer the question.
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