Chem 14BL Final Cheat Sheet|ALL YOU NEED

A−¿ ¿ ¿ ¿ (base/acid) HA=acid OH- =base A - =conjugate base H2O=conjugate acid Initial pH: Soltn. Acetic acid made by dissolving 5.00x10-3 mol acetic acid (HA) in 100.00mL water. Then titrated with strong base (NaOH) conc. 0.250M (Ka = 1.8x10-5) HA + H2O  H3O + + AKa = (x )(x) 0.05−x ASSUME X<<0.005  1.8x10-5 = (x)(x ) 0.05 solve I 0.05M ~ 0 0 x = 3.0x10-4 CHECK ASSUMPTION (3.0 x10−4 ) 0 .005 ∗100 = 6% (doesn’t work, plug into quadratic, watch signs) C -x ~ +x +x pH = -log(H3O + ) = -log(x) = -log(2.91x10-4) = 3.02 USE M FOR THIS PROBLEM NOT MOL E 0.005 – x ~ +x +x Buffer Region B4 Eq. Pt.: add 10mL NaOH mol = M*L = 0.250M * 0.01L = 0.0025mol NaOH and 0.005mol A.A. HA + OH-  H2O + ApH = pKa + log [ A] [HA ] = -log(1.8x10-5) + log 0.0025 0.0025 = -log(1.8x10-5) + 0  pH=4.74 I 0.005 mol 0.0025 ~ 0 C -0.0025 -0.0025 ~ +0.0025 E 0.0025 0 ~ 0.0025 Equivalence Point: add 20mL NaOH Reversed ICE Table (works for weak base init pH) Kw=10-14 Kw = Ka+Kb Kb= 10−14 1.8 x10−5 = 5.5555x10-10 HA + OH-  H2O + A- A- + H2O  HA + OHKb = (x )(x) 0.04166−x ASSUME x<<0.04166  x2 = (5.555x10-10)