Acid-Base Chemistry: Concepts, Calculations, and Equilibrium

Posted on Mar 27, 2025 in Chemistry

CATION

ANION

SOLUTION WILL BE

ACIDIC

NEUTRAL

ACIDIC

NEUTRAL

NEUTRAL

NEUTRAL

NEUTRAL

BASIC

BASIC

ACIDIC

BASIC

Ka > Kb ACIDIC

Ka = Kb NEUTRAL

Ka < Kb BASIC


Ka = ([H+] x [B–]) / [HB]

  1. Measure known amount of the weak acid, HB

  2. Determine [H+]eq by measuring pH if necessary

  3. Use the balanced equation to find [B–]eq

  4. Calculate [HB]eq which is [HB]o – [H+]eq

  5. Solve for Ka

Larger Ka value = the stronger the weak acid

Smaller pKa value = the stronger the weak acid

Weak Acids:

  1. Ionizable hydrogen atom

  2. Anions with an ionizable hydrogen atom

  3. Cations

    1. Protonated bases

    2. Metal cations

Weak Bases:

  1. Molecules (nitrogen compounds with unpaired electrons)

  2. Anions (that are the conjugate base of a weak acid)

  3. *********DOES NOT INCLUDE WATER*********


Ka x Kb = 1.0 x 10-14


Stronger the weak acid, the weaker the conjugate base

Stronger the weak base, the weaker the conjugate acid

The conjugate base of a weak acid is a WEAK BASE

The conjugate acid of a weak base is a WEAK ACID


Percent Ionization

[HB]eq = [HB]o – [H+]eq

% ionization = ([H+]eq / [HB]o) x 100


>5% – successive approximations are needed (where [HB]eq is the [H+] used from the previous assumption.)

  • First ionization, x (not x2) is equal to 0. Just multiply given values and obtain [H+]

  • Second ionization, the obtained [H+] from previous will be used as x

  • If percent ionization does not change to below 5% for several tries, STOP.


Ka1 > Ka2 > Ka3



NEUTRAL IONS

BASIC IONS

ACIDIC IONS

ANIONS

Cl–, Br–, I–, NO3–, ClO4–, SO42-

C2H3O2–, CO32-, F–, PO43-, NO2–, HCO3–, S2-, HS–, CN–, HPO42-

HSO4–, H2PO4–

CATIONS

Li+, Na+, K+, Ca2+, Sr2+, Ba2+

NONE

NH4+, Mg2+, Al3+


Kp – partial pressure equilibrium constant

Kc – concentration equilibrium constant

Kp = Kc (RT)n        T = Kelvins (273.15+C)       R = 0.08206


Q > K

Shifts to the left

Q = K

No shift

Q < K

Shifts to the right

Quadratic Formula – x = (-b ± √(b2-4ac)) / 2a (take the value that makes equilibriums POSITIVE)


n

P increases/V decreases

P decreases/V increases

+

LEFT

RIGHT


RIGHT

LEFT

0

NO EFFECT

NO EFFECT

ENDOTHERMIC: T increases RIGHT // T decreases LEFT

EXOTHERMIC: T increases LEFT // T decreases RIGHT

P = MRT OR P = (nRT)/V


STRONG ACIDS

HCl HBr HI HNO3 HClO4 H2SO4

STRONG BASES

LiOH NaOH KOH Ca(OH)2 Sr(OH)2 Ba(OH)2


H2O is amphiprotic

H3O+ (base) ← H2O → OH– (acid)

Bronsted-Lowry Acid (donates H+) similar compound or molecule is its conjugate base

Bronsted-Lowry Base (accepts H+) similar compound or molecule is its conjugate acid


Lewis Acid accepts electron pair

Lewis Base donates electron pair

ALL Lewis Bases ARE ALSO Bronsted-Lowry Bases

NOT ALL Lewis Acids ARE ALSO Bronsted-Lowry Acids

Lewis Acids are CATIONS (+ charged), and molecules with incomplete octets (Al3+ and BF3)


Kw = [H+] x [OH–] = 1.0 x 10-14 If one increases, the other must decrease!

Basic solutions = Alkaline solutions

pH = -log[H+] or [H+] = 10-pH

pOH = -log[OH–] or [OH–] = 10-pOH

pH + pOH = 14

pH = <7             pOH = >7 Neutral = 7

STRONG ACID IS GIVEN:

[OH–] = (1.0 x 10-14)/[H+]

STRONG BASE IS GIVEN:

[H+] = (1.0 x 10-14)/[OH–]

Red litmus – Blue = BASIC

Blue litmus – Red = ACIDIC