TOPIC 12 - ACID + BASE EQIUILIBRIA

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Total cards: 24•Content version: v4•Public deck•Updated: 8/1/2026

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Front (Question)

Bronsted-Lowry acid eq

Back (Answer)

substance that is a proton donor (H+) HA(aq) +H2O (l) –> H3O+ + A-

Front (Question)

Bronsted-Lowry Base eq

Back (Answer)

substance that can accept a proton - anything with a lone pair of electrons = dative covalent bond formed B(aq) + H2O –> BH+ + OH-

Front (Question)

strong acid vs weak acid

Back (Answer)

strong acid is able to completely dissociate. weak acid partially dissociate (very slightly) CH3COOH = CH3COO- + H+

Front (Question)

weak bases

Back (Answer)

only slightly protonate in H20 NH3(aq) = NH4+ + OH-

Front (Question)

examples of strong acids

Back (Answer)

HCl, H2SO4 (dibasic acid) , HNO3, HI, HBr, HClO, H3O+

Front (Question)

examples of weak acids

Back (Answer)

HSO4- , H3PO4 (tribasic acid), HNO2, HF, HCN, NH4+, H2S, H2CO3

Front (Question)

define pH

Back (Answer)

• log[H+] = concen of H+ in sol - always 2d.p

Front (Question)

find [H+]

Back (Answer)

10^-pH

Front (Question)

amphoteric substances

Back (Answer)

H2O & NH3 can act as an acid and base H2O + NH3 = NH4+ + H20 equil favour direction of where the stronger acid/base produces weaker acid and weaker base

#10

Front (Question)

monoprotic and diprotic acid dissociation

Back (Answer)

H2SO4 –> H+ + HSO4- HSO4- –> H+ + SO4- calc pH the slightly lower pH of first ionisation as not many H+ from HSO4- dissociate

#11

Front (Question)

for every change in 1 unit of pH the times of dilution/ concentration

Back (Answer)

x10^-1 (concen H+) = pH 1 x10^-2 = pH 2 x 10^1 (concen H+) = pH -1 but for weak acids the pH will increase less than 1 unit

#12

Front (Question)

acid + base reaction involves and eq

Back (Answer)

involves the transfer of a proton acid + base -> conjugate acid (paired with base) + conjugate base A + B –> BH+ + A- to identify need to by looking both directions of the reaction

#13

Front (Question)

conjugate acid (BH+)

Back (Answer)

base (B) accepts a proton and forms a species

#14

Front (Question)

conjugate base (A-)

Back (Answer)

acid (HA) donates the proton and species is formed

#15

Front (Question)

Stronger the acid

Back (Answer)

the weaker the conjugate base

#16

Front (Question)

weaker the acid

Back (Answer)

the stronger the conjugate acid

#17

Front (Question)

if asked to find pH when giving the concen of a base

Back (Answer)

Ph + Poh = 14 rearrange after finding Poh= -log[OH-] then find Ph = ANS e.g. 0.15 NaOH mol/dm3. find pH? [oh] = [0.15] Poh = -log[0.15] = 0.8239 = 14-0.8239 pH = 13.17

#18

Front (Question)

Ph + Poh = 14 only happens when

Back (Answer)

Kw = 1x10^-14 If Kw = 1x10^-15 then Ph + Poh = 15 etc

#19

Front (Question)

calc pH of weak acids remember [H+]*mole ratio in equation ALWAYS WRITE EQ

Back (Answer)

- partial dissociation HA = H+ + A- Ka (expresses how easily the acid releases H+) Ka = [H+] [A-]/[HA] due to 1:1 ratio Ka = [H+]^2/[HA]initial rearrange to find [H=] then pH can find pOH = 14 -pH

#20

Front (Question)

presentable/ manageable version of Ka do not deal with standard form

Back (Answer)

pKa = -logKa Ka = 10^-pKa

#21

Front (Question)

the smaller the pKa

Back (Answer)

the greater the Ka greater the dissociation the stronger the acidity strength of acid increases pH<7

#22

Front (Question)

pH of weak acids (Ka) assumptions

Back (Answer)

• [H+] = [A-] at equilibrium they have dissociated at 1:1 ratio • the amount of dissociation is small so assume inital concen of undissociated acid is CONSTANT and the acid that has been dissociated is CONSTANT = Ka

#23

Front (Question)

ionic product of water

Back (Answer)

Kc = [H+] [OH-] / [H20] Kc*[H2O] = [H+][OH-] constant as cocne on ions bigger than h2o = Kw = [H+][OH-] mol2/dm6 H20 dissociates slightly so endo so shit to right

#24

Front (Question)

25’C Kw=

Back (Answer)

1x10^-14 mol2/dm6