## Sapling #7

Moderators: Chem_Mod, Chem_Admin

805377003
Posts: 102
Joined: Wed Sep 30, 2020 10:10 pm

### Sapling #7

The Kb for an amine is 4.352×10−5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.258 ? Assume that all OH− came from the reaction of B with H2O.

Can someone explain this?

Peter DePaul 1E
Posts: 91
Joined: Wed Sep 30, 2020 9:55 pm
Been upvoted: 1 time

### Re: Sapling #7

So we're dealing with the equilibrium of $NH_{3}+H_{2}O\rightleftharpoons NH_{4}^{+}+OH^{-}$. Therefore $K_{b}=\frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]}$. We know from the pH the pOH = 14 - 9.258 = 4.742, therefore we can determine $[OH^{-}]=10^{-4.742}=1.81\times 10^{-5}M$. We also know $[NH_{4}^{+}]=[OH^{-}]$. Now we know that $[NH_{3}]=\frac{(1.81\times 10^{-5})^{2}}{4.352\times 10^{-5}}=7.53\times 10^{-6}M$. Percent protonated = $\frac{1.81\times 10^{-5}}{1.81\times 10^{-5}+7.53\times 10^{-6}}*100=70.6%$ protonated.

Return to “Equilibrium Constants & Calculating Concentrations”

### Who is online

Users browsing this forum: No registered users and 1 guest