Acids and Bases Flashcards
What is a Brønsted acid and base in terms of protons and hydrogen ions
Acid = proton donor, release H+
Base = proton acceptor, attract H+
What does Brønsted acid-base equilibria involve
Transfer of protons
Equation that related pH and H+
pH = -log10[H+]
[H+] = 10^-pH
What is Ka
An equilibrium constant for the dissociation of an acid
What is the equation for Kw in pure water
Kw = [H+]^2
Equation for pKa
pKa = -log10Ka
What does it mean when acids and bases dissociate in water
Break up into positively and negatively charged ions
What dissociates the most and why
Strong acids = releasing all H+ ions
Strong bases = ionise completely
Weaker bonds dissociate more
What happens to the equilibrium if you add more acid or base in acid + base = salt + water
Position of equilibrium shifts right
(opposite happens if you add more salt or water)
What does the dissociation of water equation look like
H2O <–> H+ + OH-
What is a conjugate acid-base pair
A conjugate acid-base pair is two species that are different from each other by an H+ ion
what is Kc
the equilibrium constant, measured using products and reactants in either the gaseous or aqueous state
What is Kw
Kw i= autoprolysis constant of water
(at 25 degrees Celsius)
Is always = 1.0 x 10^-14.
What is the Kc and Kw equation for the dissociation of water
Kc = [H+][OH-] / [H2O]
Kw = [H+][OH-]
Why is [H2O] not shown in the Kw expression
[H2O] is almost constant
What is a pH scale a measure of
Hydrogen ion concentration in a solution
measured in mol dm-3
What is the equation for strong acid ionisation
HA (aq) -> H+ (aq) + A- (aq)
What is the equivalence point
where the amount of acid neutralises the alkali or vice versa
Kc equations
Kc x H2O = [H2O][OH-]
Strong base ionisation equation
BOH (aq) → B+ (aq) + OH- (aq)
Strong base equations:
[H+] = Kw/[OH-}
[H+] = 1 x 10^-14 / [OH-]
pH = -log[H+] = -log (Kw (1 x 10^=14) / [OH-])
Calculate the pH of 0.15 mol dm-3 sodium hydroxide, NaOH
[H+] = Kw ÷ [OH-]
[H+] = (1 x 10-14) / 0.15 = 6.66 x 10-14
pH = -log[H+]
= -log 6.66 x 10-14 = 13.17
Calculate the hydroxide concentration of a solution of sodium hydroxide when the pH is 10.50
Step 1: Calculate H concentration
[H+]= 10-pH
[H+]= 10-10.50
[H+]= 3.16 x 10-11 mol dm-3
Step 2: Rearrange the ionic product of water
Kw = [H+] [OH-]
[OH-]= Kw ÷ [H+]
Step 3: Find the concentration of hydroxide ions
Kw is 1 x 10-14 mol2 dm-6,
[OH-]= (1 x 10-14) ÷ (3.16 x 10-11)
[OH-]= 3.16 x 10-4 mol dm-3
What is Ka
Dissociation constant for a weak acid