6D.5
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 112
- Joined: Sat Jul 20, 2019 12:16 am
Re: 6D.5
You need to make an ice table for the reaction of NH3 reacting with H2O. Once you solve for the equilibrium concentration of OH-, you can find pOH (pOH= -log[OH-]). Then you can convert pOH to pH (pH + pOH = 14). Percentage protonation is calculated by doing ([NH4+]/[NH3]initial) x100%
-
- Posts: 98
- Joined: Wed Sep 18, 2019 12:18 am
Re: 6D.5
Emily Chirila 2E wrote:You need to make an ice table for the reaction of NH3 reacting with H2O. Once you solve for the equilibrium concentration of OH-, you can find pOH (pOH= -log[OH-]). Then you can convert pOH to pH (pH + pOH = 14). Percentage protonation is calculated by doing ([NH4+]/[NH3]initial) x100%
How are we supposed to solve for x in the ICE table if we do not know what Kb is?
-
- Posts: 101
- Joined: Wed Sep 30, 2020 9:43 pm
Re: 6D.5
There is a table with all of the values in the textbook. Idk what the actual place it is but if you scroll up a bit you should see all of the tables with the Kb values
Return to “Calculating pH or pOH for Strong & Weak Acids & Bases”
Who is online
Users browsing this forum: No registered users and 5 guests