Any equilibrium constant (including the acid dissociation constant) is defined as being the product of the concentrations of the products (each to the power of the number of moles) over the product of the concentrations of the reactants (each to the power of the number of moles), e.g. for a reaction:

aA + bB ⇌ cC + dD

Therefore the correct answer here is:

\(K_a = {{[CH_3CO_2^-][H^+]} \over [CH_3CO_2H]}\)

Any equilibrium constant (including the base dissociation constant) is defined as being the product of the concentrations of the products (each to the power of the number of moles) over the product of the concentrations of the reactants (each to the power of the number of moles), e.g. for a reaction:

aA + bB ⇌ cC + dD

However, the concentration of water in an aqueous solution is effectively constant, so [H₂O] is incorporated into the constant. Therefore the correct answer here is:

\(K_b = {{[NH_4^+][OH^-]} \over [NH_3]}\)

Strong acids (like hydrochloric) and alkalis fully dissociate into ions in aqueous solution, weak acids (like ethanoic) and alkalis only partially dissociate.

\(K_a = {{ [A^-][H^+] } \over [HA]}\)

Therefore the K_a for a strong acid will be greater than for a weak acid (more ions on top of fraction).

pKa = −log₁₀K_a

Therefore higher K_a leads to lower pK_a (similarly to the relationship between [H⁺] and the pH scale).

Strong acids will always have lower pH than weak acids of the same concentration, again because of the dissociation.

Therefore 1 only is the correct answer.

Strong acids and bases (like sodium hydroxide) fully dissociate into ions in aqueous solution, weak acids and bases (like ammonia) only partially dissociate.

\(K_b = {{ [OH^-][BH^+] } \over [B]}\)

Therefore the K_b for a strong base will be greater than for a weak base (more ions on top of fraction).

pK_b = −log₁₀K_b

Therefore higher K_b leads to lower pK_b (similarly to the relationship between [H⁺] and the pH scale).

Strong bases will always have higher pH than weak bases of the same concentration, again because of the dissociation.

Therefore 1 and 3 only is the correct answer.

K_a = 1.78 × 10⁻⁴ (given in question).

pK_a = −log₁₀K_a = −log₁₀1.78 × 10⁻⁴ = 3.74958

For any given conjugate acid-base pair at 298K pK_a + pK_b = 14. The methanoate ion is the conjugate base of methanoic acid.

Therefore, methanoate ion pK_b = 14 − 3.74958 = 10.25042 = 10.3 (3sf)

10.3 is thus the corect answer.

(This can also be calculated using K_a × K_b = K_w)

Incorrect answers

17.8 is a missed negative sign in the pK_a expression.

10.2 is a rounding error (a 5 rounds up by convention).

17.7 assumes a missed negative sign in the pK_a expression and includes a rounding error (a 5 rounds up by convention).

For a weak acid: HA ⇌ H⁺ + A^–

The assumptions that we make when calculating the pH of weak acids (like ethanoic) include:

We assume [H⁺] = [A^–] and therefore that there is no contribution by water to [H⁺].

And we assume that [CH₃CO₂H] at equilibrium = initial [CH₃CO₂H] since there is only partial dissociation.

This gives us the expression:

\(K_a = {{ [H^+]^2 } \over [HA]}\)

We do not assume that [CH₃CO₂H] = [CH₃CO₂⁻].

Therefore 2 and 3 only is the correct answer.

For a weak acid: HA ⇌ H⁺ + A^–

The assumptions that we make when calculating the pH of weak acids include:

We assume [H⁺] = [A^–] and therefore that there is no contribution by water to [H⁺].

And we assume that [HA] at equilibrium = initial [HA] since there is only partial dissociation.

This gives us the expression:

\(K_a = {{ [H^+]^2 } \over [HA]}\)

Using the pK_a given, K_a = 10^–pKa = 10^−3.75 = 1.77828 × 10⁻⁴

And using the equation one row above: 1.77828 × 10⁻⁴ = [H⁺]² / 0.20

[H⁺]² = 1.77828 × 10⁻⁴ × 0.20 = 3.5565 × 10⁻⁵

[H⁺] = 0.0059637

pH = −log₁₀ 0.0059637 = 2.224485 = 2.2 (2sf)

Therefore 2.2 is the correct answer.

Incorrect answers

4.4 is forgetting the square root

0.062 is using pKa (3.75) in the Ka expression instead of Ka (1.77828 × 10⁻⁴)

0.70 assumes a strong acid, so pH = −log₁₀ 0.20

For a weak acid: HA ⇌ H⁺ + A^–

The assumptions that we make when calculating the pH of weak acids include:

We assume [H⁺] = [A^–] and therefore that there is no contribution by water to [H⁺].

And we assume that [HA] at equilibrium = initial [HA] since there is only partial dissociation.

This gives us the expression:

\(K_a = {{ [H^+]^2 } \over [HA]}\)

1.35 × 10⁻⁵ = [H⁺]² / 0.50

[H⁺]² = 1.35 × 10⁻⁵ × 0.50 = 6.75 × 10⁻⁶

[H⁺] = 0.0025981

pH = −log₁₀ 0.0025981 = 2.5853481 = 2.6 (2sf)

Therefore 2.6 is the correct answer.

Incorrect answers

5.2 is forgetting the square root

−2.6 is forgetting the minus in the pH expression

−5.2 is forgetting the square root and the minus in the pH expression

For a weak base: B + H₂O ⇌ BH⁺ + OH^–

The assumptions that we make when calculating the pH of weak bases include:

We assume [OH⁻] = [BH⁺] and therefore that there is no contribution by water to [OH⁻].

And we assume that [B] at equilibrium = initial [B] since there is only partial dissociation.

This gives us the expression:

\(K_b = {{ [OH^-]^2 } \over [B]}\)

4.57 × 10⁻⁴ = [OH⁻]² / 0.50

[OH⁻]² = 4.57 × 10⁻⁴ × 0.50 = 2.285 × 10⁻⁴

[OH⁻] = 0.0151162

pOH = −log₁₀ 0.0151162 = 1.8205569

pH = 14−1.8205569 = 12.179443 = 12.2 (2sf)

Therefore 12.2 is the correct answer.

Incorrect answers

10.4 is forgetting the square root

1.82 is giving the pOH instead of pH

3.64 is forgetting the square root and giving the pOH instead of pH