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12 - acid-base equilibria

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

Card preview (24)

#1
Front (Question)

brønsted-lowry acids

Back (Answer)

proton donors

#2
Front (Question)

brønsted-lowry bases

Back (Answer)

proton acceptors

#3
Front (Question)

dissociate.. strong acids and bases do what?

Back (Answer)

dissociate almost completely

#4
Front (Question)

conjugates any species that has gained a proton is..

Back (Answer)

the conjugate acid

#5
Front (Question)

conjugates any species that has lost a proton is..

Back (Answer)

the conjugate base

#6
Front (Question)

acid + base

Back (Answer)

salt + water

#7
Front (Question)

acid + metal

Back (Answer)

salt + hydrogen

#8
Front (Question)

acid + carbonate

Back (Answer)

salt + water + carbon dioxide

#9
Front (Question)

standard enthalpy change of neutralisation

Back (Answer)

enthalpy change when acid and base solutions react together to produce 1 mole of water ALWAYS EXOTHERMIC

#10
Front (Question)

[H⁺] pH

Back (Answer)

-log[H⁺]

#11
Front (Question)

ph [H⁺]

Back (Answer)

10⁻(pH)

#12
Front (Question)

(2) monoprotic acids

Back (Answer)

• dissociate to produce one H⁺ ion • conc of acid = conc of H⁺ ions

#13
Front (Question)

(2) diprotic acids

Back (Answer)

• dissociate to produce two H⁺ ions • conc of acid = 2 x conc of H⁺ ions

#14
Front (Question)

formula Kw

Back (Answer)

[H⁺][OH⁻]

#15
Front (Question)

what is Ka used for?

Back (Answer)

to find the pH of weak acids

#16
Front (Question)

(2) what are the assumptions for Ka?

Back (Answer)

• only a small amount of HA dissociates * [HA] eq ≈ [HA] start • all H⁺ ions come from the acid

#17
Front (Question)

(2) formula Ka

Back (Answer)

Ka = [H⁺][A⁻] / [HA] Ka = [H⁺]² / [HA]

#18
Front (Question)

does water dissociate weakly or strongly?

Back (Answer)

weakly H₂O ⇌ H⁺ + OH⁻

#19
Front (Question)

value of Kw

Back (Answer)

1.00 x 10⁻⁴

#20
Front (Question)

in terms of H⁺ concs what does pure water have?

Back (Answer)

equal concentrations of H⁺ and OH⁻ so Kw = [H⁺]²

#21
Front (Question)

Kw pKw

Back (Answer)

-log [Kw]

#22
Front (Question)

pKw Kw

Back (Answer)

10⁻ (pKw)

#23
Front (Question)

Ka pKa

Back (Answer)

-log [Ka]

#24
Front (Question)

pKa Ka

Back (Answer)

10⁻ (pKa)