O ijklm: Acid-base equilibria Flashcards
Bronsted-Lowry theory
Define:
- Acids
- Bases
- Acids = proton donors
- Bases = proton acceptors
Write the formula for the conjugate base of ethanoic acid.
CH_{3}COO^{-}
Write the formula for the conjugate base of:
- Hydrochloric acid
- Chloric(VII) acid
- Sulfuric acid
- Hydrogen sulfide
- Water
- HCl → Cl^{-}
- HClO_{4} →ClO_{4}^{-}
- H_{2}SO_{4} →HSO_{4}^{-}
- H_{2}S → SH^{-}
- H_{2}O → OH^{-}
Water may act as an acid or base. Write 2 equations, including hydrochloric acid and ammonia, to demonstrate this.
Acting as a base: accepting a proton from HCl
H_{2}O_{(l)} + HCl_{(aq) }→ H_{3}O^{+}_{(aq)}_{ }+ Cl^{-}_{(aq)}
Acting as an acid: donating a proton to ammonia
H_{2}O_{(l)} + NH_{3}_{(aq) }→ OH^{-}_{(aq)} + NH_{4}^{+}_{(aq)}
- What equation is used to calculate pH?
- How can [H^{+}] be calculated from a pH value?
pH = -log_{10}[H^{+}]
[H^{+}]^{ }= 10^{-pH}
What is the [H^{+}] in water?
pH = 7
-log[H^{+}] = 7
[H^{+}] = 10^{-7} mol dm^{-3}
What can be said of [H^{+}] in the case of a solution with pH < 0?
pH < 0
-log[H^{+}] < 0
log[H^{+}] > 0
[H^{+}] > 10^{0}
[H^{+}] > 1 mol dm^{3}
Compare strong and weak acids. Use equations to demonstrate the comparison.
Strong acids dissociate protons completely in aqueous solution:
HA + H_{2}O → H_{3}O^{+} + A^{-}
HA → H^{+} + A^{-}
Weak acids do not dissociate protons completely in aqueous solution:
HA + H_{2}O ⇌ H_{3}O^{+} + A^{-}
HA ⇌ H^{+} + A^{-}
The further to the left the eq. position lies, the weaker the acid
Write equations to demonstrate why solutions of carbon dioxide are acidic.
CO_{2 (aq) }+ H_{2}O ⇌ H^{+}_{(aq)} + HCO_{3}^{-}_{(aq)}
HCO_{3}^{-}_{(aq)} ⇌ H^{+}_{(aq)} + CO_{3}^{2-}_{(aq)}
Weak acidity; equilibrium lies to the left.
- Write an equation for the equilibrium constant of a weak acid, used to find [H^{+}].
- Name this constant
HA ⇌ H^{+}^{ }+ A^{-}
K_{a}_{ }= [H^{+}] [A^{-}] / [HA]
K_{a}, acidity constant / acid dissociation constant
To find [H^{+}] for a weak acid by calculating K_{a}, what 2 assumptions are made for simplicity?
1: [H^{+}] = [A^{-}]
Negate small number of protons added by ionisation of water: H_{2}O ⇌ H^{+} + OH^{-}
2: [HA_{eq}] = [HA_{initial}]
Negate fraction of HA which have dissociated H^{+} - small since weak acid
To find [H^{+}] for a weak acid by calculating K_{a}, the following assumptions are made. When are they not good assumptions?
1: [H^{+}] = [A^{-}] Negate small number of protons added by ionisation of water: H_{2}O ⇌ H^{+} + OH^{-}
2: [HA_{eq}] = [HA_{initial}] Negate fraction of HA which have dissociated H^{+}
- For very dilute solutions
- For stronger acids
Calculate the pH of a solution of methanoic acid made up by adding 0.25 mol to 100 cm^{3} of water. K_{a} at 298 K = 1.6 x 10^{-4} mol dm^{-3}.
K_{a}_{ }= [H^{+}]^{2} / [HA]
[H^{+}]^{2} / (0.25/0.1) = 1.6 x 10^{-4}
[H^{+}] = √(1.6 x 10^{-4} x 2.5)
pH = -log(√(1.6 x 10^{-4} x 2.5)) = 1.7
How do values of K_{a} compare for strong and weak acids?
For stronger acids, [HA] is lower and [H^{+}] is greater, so K_{a}_{ }is greater.
Since values of K_{a} for weak acids can be small, a log scale pK_{a} is often used. How is pK_{a} calculated?
pK_{a} = -log_{10}(K_{a})
The weaker the acid, the __ the value of pK_{a}
Greater