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chemistry acid and base

calculate [OH- ] and pH for each of the following solutions. (Kw = 1.0 10-14.)

A mixture that is 0.050 M in CH3COONa and 0.044 M in (CH3COO)2Ba

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Stanton D. | Tutor to Pique Your Sciences InterestTutor to Pique Your Sciences Interest
4.6 4.6 (42 lesson ratings) (42)
This is a standard acid/base problem. You need to
1) add up all the acetate ion concentrations (as initially dissolved, before anything else reacts with them!)
2) Recognize that acetate is the conjugate base for acetic acid; some of the acetate will pick up protons from the water to become acetic acid; (these protons are NOT acquired (mostly) from the 10^-7 M amount of initial H+ in pure water! instead, let OH- be formed)
3) Write equations relating the acetic acid deprotonation equilibrium and water dissociation equilibrium
4) Combine the K's for the two reactions (multiply the equilibrium concentration expressions == multiply the K's) -- you are trying to remove OH-, and get an expression with H+ in instead
5) Set up an "ICE" table for acetate, acetic acid, and H+ (ICE= initial, change, equilibrium) molar concentrations, using "x" as unknown for the molarity of acetic acid formed 
6) Solve the quadratic expression you obtain for the value of "x"
7) Plug back in to calculate whatever you don't have yet.