Sapling Week 2 (Question 5)

[Megan Verschuur 1C]

I am struggling with question 5 on sapling week 2. Would someone be able to tell me if I am going about this problem the right way or if I need to modify my approach?

1.To find the pOH, I did 14-9.025= 4.975
2. 4.975=-log[OH-], which I got 1.1 * 10^-5 = [OH-]
3. According to the hint given, [OH-]=[BH+] which would both be (1.1 * 10^-5)-x, but since Kb is so small, we can approximate.
4. (6.108*10^-5)= (1.1 * 10^-5)^2 / (1.21*10^-10)

Following these steps I got a negative protonation percentage, which does not make sense. Help! Thank you in advance!

[Alison Perkins 2B]

I wouldn't approximate, but rather set up an ICE table using another variable as the initial concentration. If you do 10^(-pOH) you will get an x value for your table. Then you can use the x and Kb to solve for an initial concentration, which you can use to set up the percent ionization equation. I hope that helps!