15&16+Equilibrium+Acid-Base

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Notes:
Conjugate Acid/Base Explained in words: http://www.science.uwaterloo.ca/~cchieh/cact/c123/baseacid.html

Notes on finding the pH of salts:
**Problem #1:** The Ka of formic acid (HCOOH) is 1.8 x 10¯4. What is the pH of 0.35 M solution of sodium formate (NaHCOO)? **Solution:** Calculate the Kb of the formate (HCOO¯) ion: > 1.00 x 10¯14 = (1.8 x 10¯4) (x)x = 1.00 x 10¯14 / 1.8 x 10¯4 = 5.56 x 10¯11 Calculate the [OH¯]: > 5.56 x 10¯11 = [(x) (x)] / (0.35 - x)neglect the minus x > x = 4.41 x 10¯6 M Calculate pOH, then pH: > pOH = - log 4.41 x 10¯6 = 5.36pH = 14 - pOH = 14 - 5.36 = 8.64

**Problem #2:** Calculate the pH of a 0.36 M CH3COONa solution given the pKa of acetic acid is 4.745. **Solution:** Using the pKa, calculate the Ka of acetic acid: > Ka = 10¯pKa = 10¯4.745 = 1.80 x 10¯5 Using the Ka of the acid, find the Kb of the base (the CH3COONa): > 1.00 x 10¯14 = (1.80 x 10¯5) (x)x = 5.56 x 10¯10 Use the Kb expression to calculate the [OH¯]: > 5.56 x 10¯10 = [(x) (x)] / (0.36 - x)neglect the minus x > x = 1.415 x 10¯5 M (I threw in a guard digit.) Calculate pOH, then pH: > pOH = - log 1.415 x 10¯5 = 4.85pH = 14 - pOH = 14 - 4.85 = 9.15 Comment on the solution to Example #3: You can use the pKa to get the pKb > 4.745 + x = 14.000x = 9.255 Use the pKb to get the Kb: > Kb = 10¯pKb = 10¯9.255 = 5.56 x 10¯10 Go to rest of solution, at "Use the Kb expression," just above.

**Problem #3:** What is the pH of a 0.0510 molar solution of salt NaCN (the Ka for HCN is 6.166 x 10¯10)? **Solution:** CN¯ is the salt of a weak acid and hydrolyzes thusly: > CN¯ + H2O ⇔ HCN + OH¯ Determine the Kb for this reaction (using Kw = KaKb): > 1.00 x 10¯14 = (6.166 x 10¯10) (x)x = 1.6218 x 10¯5 (I kept a couple guard digits.) Write the Kb expression and solve for [OH-]: > Kb = ([HCN] [OH¯)] / [CN¯]1.6218 x 10¯5 = [(x) (x)] / 0.0510 > x = 9.0946 x 10¯4 M Determine pOH (with negative log) and then pH (using pH + pOH = 14): > -log 9.0946 x 10¯4 = 3.0412pH = 14 - 3.0412 = 10.959 (to three sig figs)

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