Acid-Base Equilibrium

Question 1

Acids

Answer 1

Proton donor, must have an H+ to give up

Question 2

Bases

Answer 2

Proton acceptor, must have a lone pair of e- that can form a coordinate covalent bond (CCB) with H+ OR a negative charge to attract an H+ (F-)

Question 3

Acid Base Reaction

Answer 3

Involves transfer of H+ from acid to base (Competition for a p+ between two bases)

Question 4

Salt

Answer 4

Substance that can produce an acidic or basic solution

Question 5

Amphoteric Substance and Hydrolysis of Amphoteric Ions

Answer 5

A substance that can act as a B-L acid or base. Has an H+ to donate and a -charge / lone pair to accept a proton
All polyatomic ions whose chemical formula begins with H are amphoteric (HC3-), these can either donate or receive a proton when dissolved in water.

Question 6

Conjugate Acid-Base Pair

Answer 6

Two substances whose chemical formula differs by only one H+

Question 7

[H3O+] and pH

Answer 7

pH = -log[H3O+] or [H3O+] = 10^-pH

Question 8

[OH-] and pOH

Answer 8

pOH = -log [OH-] or [OH-] = 10^-pOH

Question 9

Eq law for the rxn of H2O undergoing autoionization

Answer 9

K(w) = [H3O+] [OH-]

Question 10

[H3O+] vs [OH-]

Answer 10

Acid Solutions: Dissolving acids in H2O : [H3O+] >1.0x10^-7M, [H3O+] >[OH-]
Basic Solutions: Dissolving bases in H2O : [OH-] >1.0x10^-7M, [H3O+] < [OH-]
Neutral : =

Question 11

Percent Ionization

Answer 11

Indicates the extent to which an acid will ionize in water (SA have 100 percent)
p = ( [ionized acid] / original [acid solute] )x 100
p = ( [H3O+] / [weak acid] )x 100

Question 12

Ionization Constants

Answer 12

Weak Acid Ionization constant Ka = [H3O+] [acid ion] / [HA]

Weak Bases Ionization constant Kb = [OH-] [HB] / [base ion]

Question 13

Polyprotic Acids

Answer 13

Possess more than one ionizable proton (H+), donate in steps, Ka decreases per step so that K1>K2>K3…

Question 14

Hydrolysis

Answer 14

The reaction of an ion with water to produce an acidic or basic solution, ion exchanges a H+ with a water molecule in a B-L acid-base rxn

Question 15

Salts that form Neutral Solutions

Answer 15

Salt contains cation of a strong base (group 1/2) AND anions of strong monoprotic acids (Cl-, Br-, I-, NO3-)
They do not exchange an H+ with water and thus have no effect on pH of solution

Question 16

Salts that form Acidic Solutions

Answer 16

Salt contains cations that are weak acids

Question 17

Salts that form Basic Solutions

Answer 17

Salt contains anions that are weak bases

Question 18

Salts that act as Acid and Base

Answer 18

Salts contain a weak base and a weak acid, may form an acidic/basic solution.

Question 19

Strong vs Weak Acids

Answer 19

SAs completely ionize in H2O, produce high [H3O+], great electrolytes, undergo rapid rxns
WAs partially ionize in water b/c of strong attraction for their H+, creates an easy system that lies far to the left, poor electrolytes, undergo slow rxns

Question 20

Strong Bases vs Weak Bases

Answer 20

SBs completely dissociate in H2O, produce metal cations and OH- ions, great electrolytes, undergo rapid rxns
WBs don’t ionize completely in H2O, are B-L base that have poor attraction for H+, usually non-metal/polyatomic anion that reacts w/H2O to form eq that includes OH- : B- + H2O HB + OH-, poor electrolytes

Question 21

Buffers

Answer 21

Resists large changes in pH when moderate amount of acid/base is added by shifting in the direction that opposes the stress acc to LeChat’s Principle.

Question 22

Buffer Capacity

Answer 22

Buffer Capacity is the limit to the amount of SA/SB that can be neutr. by buffer before pH begins to rise rapidly. Is determined by the [ ] of the conjugate A-B pair. Eq for determining pH of Buffer: [H3O+] = x = ka( [HA]/[A-] ); [OH-] = x = kb( [HB]/[B-] )

Question 23

SA-SB Titration Calculation

Answer 23

@beginning : 100 percent ionzes, pH = -log[H3O+]
*b/c H+ = H3O+
After add: NIE, ICE, pH = -log( )
@eq : pH = 7

Question 24

WA-SB Titration Calculation

Answer 24

@beginning : x = [H3O+] = sqrt. Ka[HA]
After add : x = [””] = Ka( [HA]/[A-] ) assumptions
@eq : x = [OH-] = sqrt. Kb[B-]
Beyond : HA + OH- -> H2O + A-, ICE, pOH = -log, pH = 14-pOH