Homework 2 Question 5

[Taylor Vargas]

Hi can anyone help with question 5 on homework 2. Ive tried all the advice on here but keep getting the wrong answer. I don't know what Im doing wrong. Thank You

The Kb for an amine is 4.671×10−5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.632 ? Assume that all OH− came from the reaction of B with H2O.

[chantalaguilar3C]

You have to work backwards in a way, with your given pH you will find the pOH which would be pOH=14-9.632 which is 4.368, with that you will now find the concentration of OH- using the equation of [OH-]=10^-pOH which gives you 4.29x10^-5 so with that you will now set up the equilibrium constant equation using your given Kb so it should look like 4.671x10^-5=(4.29x10^-5)^2/x-4.29x10^-5. Now you solve for x and once you find it you will set up the equation for percentage pronated which is (4.29x10^-5/(what x is)) x 100. The result will be the percentage you are looking for.

[Katelynn Pham 1D]

Knowing that the Kb = 4.671 x 10^-5 and the pH of the solution is 9.632, you would first have to find the pOH.

pOH = 14.0 - 9.632 = 4.368

Find the molar concentration of OH-.
[OH-] = 10^-4.368 = 4.2855 x 10^-5

Using an ICE table, you can find that…
Kb = 4.671 x 10^-5 = ((4.2855 x 10^-5)^2) / (x - 4.2855 x 10^-5)

After you have found x using the quadratic equation, find the percent protonation by taking the molar concentration of OH- and dividing it by the x value, then multiplying it by 100%. Hope this helped!