(KEY)
WS Acid-Base Review
(Ka, Kb)
(KEY)
Ba(OH)2 + 2 HBr 2 H2O + BaBr2
0.0500 L Ba(OH)2 x 0.140 mol Ba(OH)2 x 2 mol HBr x 1 L HBr =
1 L Ba(OH)2
1 mol Ba(OH)2 0.100 mol HBr
0.100 x 0.140 = 0.140 L = 140 mL
0.100
CH3COOH + NaOH H2O + CH3COONa
0.04600 L NaOH x 0.50 mol NaOH x 1 mol CH3COOH = 0.023 mol CH3COOH
1 L NaOH
1 mol NaOH
0.023 mol = 2.3 M
0.0100 L
H2C3H2O4
HC3H2O4
–
H2C3H2O4
H+ + HC3H2O4
H+ + C3H2O4
–
2–
2 H+ + C3H2O4
Ka1 = 1.5 x 10–3
Ka2 = 2.0 x 10–6
2–
Ka = (1.5 x 10–3)( 2.0 x 10–6)
Ka = 3.0 x 10–9
“highest pH” means “most basic” or more [OH–] or less [H+]
Anions are basic (accept H+’s from H2O)
CO3
2–
+ 2 H2O H2CO3 + 2 OH–
(Six anions of strong acids CANNOT accept H+)
Which of the following is the conjugate acid of the bicarbonate ion?
(A) CO32–
The conjugate acid is formed when the
(B) C2O42–
+
–
_D_
(C) HCO3
(D) H2CO3
(E) HCO2
ion acts as a base by accepting an H .
–
HCO3 + H+ H2CO3
H3BO3  H+ + H2BO3
I 1 x 10–3 M
0M
0M
C –x
+x
+x
E 1 x 10–3 – x
x
x
–3
≈ 1 x 10 M
Ka = 6 x 10–10
6 x 10–10 =
x2
1 x 10–3
6 x 10–13 = x2
x = (6 x 10–13)^1/2 M H+
.
then, –log [H+] ≈ 6 – 7
1) rinse buret with TITRANT (not DI H2O)
2) fill buret with titrant
3) run out air bubbles
4) record initial volume of titrant
5) add analyte & indicator to flask
6) titrate to equivalence while
swirling flask to mix/react
completely
(rinsing sides of flask with H2O is
fine to wash down any titrant sides)
7) record final volume of titrant
NaF has Na+ ions and F– ions
Na+ cannot react with H2O
(cannot take OH– and release H+ to be acidic)
F– CAN react with H2O by hydrolysis rxn
(take H+ and create OH– to be basic)
F– + H2O HF + OH–
(Six anions of strong acids CANNOT accept H+)
(NO3– , SO42– , Cl– , Br– , I– , ClO4–)
0.01600 L NaOH x 0.150 mol NaOH = 0.00240 mol NaOH
1 L NaOH
HOCl
+ NaOH I
?
0.00240 mol
C –0.00240 –0.00240
E
?
0 mol
H2O + OCl–
0
0
+0.00240
0.00240 mol OCl–
One point for the correct
moles of OCl–
Propanoic acid, C2H5COOH, dissociates in water in accordance to the above
chemical equation. A 45.0 mL sample 0.250 M C2H5COOH was titrated with
0.300 M NaOH.
0.300