IB DP Chemistry Reactivity 3.1 Proton transfer reactions IB Style Question Bank HL Paper 1


Which products are formed from the neutralization of nitric acid by calcium hydroxide?
A. Calcium oxide and ammonia
B. Calcium nitrate and water
C. Calcium nitrate and ammonia
D. Calcium nitrate and hydrogen


Markscheme: B

The neutralization reaction between nitric acid (\(HNO_3\)) and calcium hydroxide (\(Ca(OH)_2\)) forms water and a salt. The balanced chemical equation for this reaction is:

\[2HNO_3(aq) + Ca(OH)_2(aq) \rightarrow Ca(NO_3)_2(aq) + 2H_2O(l)\]

Therefore, the correct answer is:

B. Calcium nitrate and water


Which combination describes a strong Brønsted–Lowry acid?


Markscheme: B

A strong Brønsted–Lowry acid is a substance that easily donates protons (H⁺ ions). Its conjugate base is weak because it doesn’t readily accept protons. So, for a strong Brønsted–Lowry acid:

1. Proton Donor: The strong acid easily donates protons. Therefore, the proton donor is the strong acid.

2. Conjugate Base: The conjugate base is formed when the strong acid donates a proton. Since the acid is strong, its conjugate base is weak. Strong acids typically have weak conjugate bases.

In general terms, a strong Brønsted–Lowry acid can be represented as follows:

Proton Donor: Strong Acid
Conjugate Base: Weak Base


What is the relationship between acid and base dissociation constants in a conjugate acid–base pair?

A. \(Ka \times Kb = Lw\)

B. \(\frac{Ka}{Kb} = Kw\)

C. \(pKa \times pKb = pKw\)

D. \(\frac{pKa}{pKb}=pKw\)


Markscheme: A

The relationship between acid dissociation constant (\(K_a\)) and base dissociation constant (\(K_b\)) for a conjugate acid–base pair can be expressed by the following equation:

\[K_a \times K_b = K_w\]

\(K_a\) is the acid dissociation constant for the acid.
\(K_b\) is the base dissociation constant for the base.
\(K_w\) is the ion product constant for water (\(1.0 \times 10^{-14}\) at 25°C).

So, the correct answer is:

A. \(K_a \times K_b = K_w\)


What is the pH at the equivalence point in this titration?

A. 8.5
B. 7.0
C. 5.0
D. 1.5


Markscheme: C


What is the order of increasing conductivity for aqueous solutions of these acids and bases at equal concentrations?

A. methylamine < ethanol < phenylamine
B. ethanol < phenylamine < methylamine
C. methylamine < phenylamine < ethanol
D. ethanol < methylamine < phenylamine


Markscheme: B

The order of increasing conductivity for aqueous solutions can be correlated with the strength of the acids or bases in those solutions. Stronger acids and bases tend to dissociate more in water, producing more ions and increasing the conductivity of the solution.

In the given table, we have \(\mathrm{p} K_{\mathrm{b}}\) values for different substances. The higher the \(\mathrm{p} K_{\mathrm{b}}\) value, the weaker the base.

Let’s analyze the substances:

1. Methylamine (\(\mathrm{p} K_{\mathrm{b}} = 3.34\)): Methylamine is a stronger base as compared to the other substances because of its lower \(\mathrm{p} K_{\mathrm{b}}\) value.

2. Ethanol (\(\mathrm{p} K_{\mathrm{b}} = 15.5\)): Ethanol is not a base; it is an alcohol. It does not undergo significant ionization in water, and hence, it does not contribute to the conductivity of the solution.

3. Phenylamine (\(\mathrm{p} K_{\mathrm{b}} = 9.13\)): Phenylamine is a weaker base compared to methylamine but stronger than ethanol.

Based on the strength of the bases (and their contribution to conductivity), the correct order of increasing conductivity is:

\[ \text{Ethanol} < \text{Phenylamine} < \text{Methylamine} \]

So, the correct answer is B. ethanol \(<\) phenylamine \(<\) methylamine.

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