Homework 2 exercise 10

[205984433]

When the pH of the solution is equal to the pKa of the conjugate acid (BH+), there are equal amounts of the weak base (B) and the the conjugate acid (BH+) in solution.

At pH values below the pKa, the charged, protonated species of BH+ will become more predominant in solution. At pH values above the pKa, the neutral, deprotonated species of B will become more predominant in solution.
pH 6.28 is less than the pKa. Therefore, the predominant species is the charged, protonated species of BH+ of the weak base.

This is the solution, but i don’t understand the reasoning behind it. Can someone please explain it to me?

[Nathan Chu 2G]

Hello,

Dr. Lavelle posted a document on his main website going over predominant species. The homework problem's solution reasons the "Chem 14A way." In Chem 14B, we use an equation (which we didn't learn) called Henderson-Hasselbach:

pH = pKa + log([A-]/[HA]) and pOH = pKb + log([HB+]/[B])

For problem ten, you need to first convert the pH to pOH and the pKa to a pKb value as the reactant contains a base. From there, you would plug the pOH and pKb into pOH = pKb + log([HB+]/[B]). Subtract pKb from both sides and you would get (some value)=log([HB+]/[B]). Get rid of the logarithm by doing 10^(some value) = 10^log([HB+]/[B]). After finishing this step, you now have a ratio. If 10^(some value) is small, then that must mean the denominator ([B]) is predominant. If 10^(some value) is large, then that must mean the numerator ([HB+]) is predominant. The acidity is determined by what the predominant species turns out to be; if HB+ is predominant, the solution will be acidic as there will be a lot of the conjugate acid. If B is predominant, the solution will be basic as there is a lot of the base.

Similar logic can be applied to question 9.

Hope this helps. Cheers!