A
CH3COO-; Acetate
Name a weak acid, ANY weak acid, and its conjugate base
examples:
acetic acid -> acetate
formic acid -> formate
phosphoric acid -> dihydrogen phosphate
water -> hydroxide
A stronger acid/base will have a ________ Ka or Kb value respectively
HIGHER/GREATER/LARGER/INCREASED/etc...
The conjugate acid of hydrogen sulfate is ______
The conjugate base of hydrogen sulfate is ______
CA: Sulfuric Acid H2SO4
CB: Sulfate SO42-
The formula that relates pH and pOH
14 = pH + pOH
A _________ is a substance that produced hydroxide ions in water
Arrhenius Base
The formula for pH, given proton concentration, is ________
A ________ is a substance that accepts a lone pair of electrons
Lewis Acid
A _______ pKa indicates a stronger acid
lower
A ________ is a substance that donates protons
Bronsted-Lowry Acid
(Not Arrhenius, as that is specifically in WATER)
The formula that relates Ka and Kb
Kw = (Ka)(Kb)
Hydroxides made of alkali and alkaline earth metals are strong bases except for those made with ____ and ____
Beryllium and Magnesium
NaCH3COO is a(n) ________ salt
Basic
Na came from NaOH (strong base)
CH3COO came from CH3COOH (weak acid)
The pH of 0.005 M HCl
-log(0.005)
pH = 2.3
50 mL of 0.5 acetic acid is titrated with 0.125 sodium hydroxide. The Kb of acetic acid is 5.6 x 10-10. What is the pH at half-equivalence?
pH = pKa (at half equivalence)
pH = -log(Ka) = -log(Kw/Kb)
pH = -log(1.8 x 10-5) = 4.74
Kb = Kw/Ka = 4.4 x 10-4
(ICE TABLE, then...)
x = sqrt(0.125*4.4*10-4)
x = [OH-] = 0.00742
pH = 14-pOH = 14+log(0.00742) = 11.87
0.03 M Phosphoric acid (first proton pKa is 2.13) is placed into water. The percent ionization is _____
H3PO4 -> H+ + H2PO4-
Ka = 10-pKa = 7.41 x 10-3
7.41 x 10-3 = ([H+][H2PO4-)/([H3PO4])
= (x2)/(0.03-x)
x = 0.0115
% = (0.0115/0.03) x 100 = 38.3% ionization
I have a beaker with 0.2 M ammonium and 0.150 M ammonia (Kb of ammonia is 1.8 x 10-5). The pH of this system is _______
Hint: There is an easy way, and there is a hard way (ICE table is the hard way)
pH = pKa + log(A-/HA)
= -log(Kw/Kb) + log(0.15/0.2)
= 9.25 - 0.125
pH = 9.12
The seven strong acids are:
Hydroiodic: HI
Hydrobromic: HBr
Hydrochloric: HCl
Perchloric: HClO4
Chloric: HClO3
Nitric: HNO3
Sulfuric (first proton): H2SO4
Mg3(PO4)2 is a _______ salt
HINT:
Mg(OH)2 has a pKb of 10.91
H3PO4 has a pKa of 2.15
Acidic salt
Phosphoric acid is a stronger acid than magnesium hydroxide is a base
The acid is stronger than the base
A buffer with a Ka of 1.202 x 10-5 has a pH of 4.6. The buffer ratio is...
pH = pKa + log(A-/HA)
4.6 = -log(1.202 x 10-5) + log(buffer ratio)
4.6 = 4.92 + log(buffer ratio)
-0.32 = log(buffer ratio)
buffer ratio = 10-0.32
Ratio = 0.4786
A buffer has a pH of 11.2, and contains 0.51 M of the conjugate acid and 0.49 M of the conjugate base. What is the Ka of the buffer?
pH = pKa + log(CB/CA)
11.2 = -log(Ka) + log(0.49/0.51)
11.217374 = -log(Ka)
Ka = 10^(-11.217374)
Ka = 6.06 x 10-12
This value is the pH of a solution where [H+] is 10,000 times less than [OH-]
[H+] = [OH-]/10,000
Kw = [H+][OH-]
Kw = ([OH-]/10,000)*([OH-])
[OH-] = sqrt(Kw * 10,000) = 0.00001
pOH = -log(0.00001) = 5
pH = 14 - pOH = 9
My buffer is 1 L, and has 0.15 M acetate and 0.2M acetic acid (Ka = 1.8 x 10-5). I drop in 0.04 moles of hydrochloric acid. What is the new pH?
CH3COO- (aq) + H3O+ (aq) --> CH3COOH (aq) + H2O
Moles of acetate: 0.150-0.04 = 0.11 moles = 0.11M
Moles of acetic acid: 0.2 + 0.04 = 0.24 moles = 0.24M
Hydrochloric acid: 0.04 - 0.04 = 0 moles
pH = pKa + log(A-/HA)
pH = -log(1.8x10-5)+log(0.11/0.24)
pH = 4.4
0.5 M sulfuric acid is placed in water (Ka of second proton is 1.2 x 10-2). What is the equilibrium pH?
Note: Consider the dissociation of the first AND second proton
x = 0.012 = [H+] after second dissociation
First + second dissociations = 0.5 + 0.012 = 0.512
pH = -log(0.512) = 0.291