IB DP Chemistry - Reactivity 3.1 Proton transfer reactions - IB Style Questions For HL Paper 1A -FA 2025

IBDP Chemistry HL Paper 1A & 1B - All Chapters

Question 

What is the concentration of \( \mathrm{OH^- (aq)} \), in \( \mathrm{mol\,dm^{-3}} \), in a solution at \(298.15\,\mathrm{K}\) with a pH of \(4.50\)?

\( \mathrm{pH=-\log_{10}[H^+]} \qquad \mathrm{[H^+]=10^{-pH}} \)

\( \mathrm{pOH=-\log_{10}[OH^-]} \qquad \mathrm{[OH^-]=10^{-pOH}} \)

\( \mathrm{K_w=[H^+][OH^-]} \)

(A) \(3.16\times10^{-10}\)
(B) \(3.16\times10^{-9}\)
(C) \(3.16\times10^{-5}\)
(D) \(3.16\times10^{-4}\)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{A}} \)

At \(298.15\,\mathrm{K}\), \(K_w=1.00\times10^{-14}\).

First calculate the hydrogen ion concentration:

\( \mathrm{[H^+]=10^{-4.50}=3.16\times10^{-5}\,\mathrm{mol\,dm^{-3}}} \)

Using:

\( \mathrm{K_w=[H^+][OH^-]} \)

\( \mathrm{[OH^-]=\dfrac{1.00\times10^{-14}}{3.16\times10^{-5}}} \)

\( \mathrm{[OH^-]=3.16\times10^{-10}\,\mathrm{mol\,dm^{-3}}} \)

Therefore, the correct answer is (A).

Question 

Which aqueous solution has the lowest pH?

(A) \(1.0\,\mathrm{mol\,dm^{-3}}\) \( \mathrm{HNO_3} \)
(B) \(2.0\,\mathrm{mol\,dm^{-3}}\) \( \mathrm{HCl} \)
(C) \(1.0\,\mathrm{mol\,dm^{-3}}\) \( \mathrm{CH_3COOH} \)
(D) \(2.0\,\mathrm{mol\,dm^{-3}}\) \( \mathrm{NaOH} \)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{B}} \)

\( \mathrm{HCl} \) and \( \mathrm{HNO_3} \) are strong acids and therefore dissociate essentially completely in aqueous solution.

For \(2.0\,\mathrm{mol\,dm^{-3}}\) \( \mathrm{HCl} \):

\( \mathrm{[H^+]\approx2.0\,\mathrm{mol\,dm^{-3}}} \)

\( \mathrm{pH=-\log_{10}(2.0)\approx-0.30} \)

The \(1.0\,\mathrm{mol\,dm^{-3}}\) \( \mathrm{HNO_3} \) solution has \( \mathrm{pH}\approx0 \), while ethanoic acid is a weak acid and sodium hydroxide is basic.

Therefore, the lowest pH is obtained with \(2.0\,\mathrm{mol\,dm^{-3}}\) \( \mathrm{HCl} \), so the correct answer is (B).

Question

What is the relationship between acid and base dissociation constants in a conjugate acid–base pair?
(A) \(K_a \times K_b = K_w\)
(B) \(\frac{K_a}{K_b} = K_w\)
(C) \(pK_a \times pK_b = pK_w\)
(D) \(\frac{pK_a}{pK_b} = pK_w\)
▶️ Answer/Explanation
Detailed solution

For any conjugate acid–base pair, the acid dissociation constant \((K_a)\) and base dissociation constant \((K_b)\) are related through the ion-product constant of water, \(K_w\).

The relationship is:
\[K_a \times K_b = K_w\]

Where:
• K_a = acid dissociation constant
• K_b = base dissociation constant
• K_w = \(1.0 \times 10^{-14}\) at 25°C

Therefore, the correct answer is:
✅ Answer: (A)

More DP Chemistry HL Paper 1 Questions..
Scroll to Top