CHEM 14B Lecture Notes - Lecture 4: Equilibrium Constant, Amphoterism, Cubic Function

• School: UCLA
• Department: Chemistry and Biochemistry
• Course: CHEM 14B
• Professor: Lavelle, L.

Document Summary

0. 482 mol n2 mol and 0. 933 mol o2 are placed in a. 10. 0l rxn vessel and form n2o, kc = 2. 0 x 10^-37. What is the composition of the equilibrium mixture. Initial [n2] = 0. 482 mol / 10. 0 l = 0. 0482 m. Initial [o2] = 0. 933 mol / 10. 0 l = 0. 0933 m. Assume that o2 changes its molar concentration by. = (2x)^2 / (0. 0482 - 2x)^2 (0. 933 - x) Since kc is very small, we can approximate. 0. 0482 - 2x = 0. 0382 and 0. 0933 - x = 0. 0933. Therefore, kc = 4x^2 / 0. 0482^2 * 0. 0933 = [o2] = 0. 0933 - x = 0. 033 m. [n2o] = 2x = 6. 6 x 10^-21. Reaction between acid and base is neutralization. Autoprotolysis : proton transfer between same type of. Equilibrium constant for autoprotolysis of water: kc = molecule. Ba(oh)2 is a strong base (completely ionized) Therefore, 0. 0030 m ba(oh)2 gives 0. 0060 m oh-