Achieve #5

[Brynn 2F]

Can someone explain why we add the [BH+] concentration to [B] when calculating the percentage protected? Full Equation is : [OH-] / [BH+] + [B] x 100%.

[Praveena Ratnavel - 1A]

I believe we are essentially solving with protonated/unprotonated * 100, and since both [BH+] and [B] are unprotonated, we would add them together.

[Alyssa Ly 2G]

Hello!

The reason why the [BH+] concentration is added to the [B] is because you need the initial concentration of B in order to calculate the percentage protonated.

After using Kb in order to calculate the [B] at equilibrium, you have to take into account the amount of [B] that was lost during the reaction to equilibrium in order to get the initial concentration of [B]. To do this, you know that x amount is lost from the initial [B] at equilibrium. As a result, this means that to get the initial [B], you have to do:

initial [B] = equilibrium [B] + x

Since [BH+] is equal to x, you would add that amount back to [B] at equilibrium in order to get the initial [B].

Remember, % protonated = [BH+]/[B]initial

Hope this is helpful! :))

[Lily Rivas 1H]

Just a further note here !

Achieve states that BH⁺ acts as the portion of the amine that is protonated. B acts as the portion of the amine that is not protonated.