NESA Chemistry Using Brønsted–Lowry Theory

This is incorrect because a lower pH value indicates a solution is more acidic, not less. Additionally, the pH scale is logarithmic, so the concentration difference is exponential ($10^{\Delta \text{pH}}$), not a linear ratio like 0.5.

While Solution A is more acidic, the concentration difference is not a factor of 2. Since the pH scale is logarithmic, a difference of 3 pH units implies a $10^3$ difference in $[H^+]$, not a factor derived from dividing the pH values ($6/3$).

Solution A has a lower pH (3) than Solution B (6), which makes it more acidic. Therefore, it must have a higher concentration of hydrogen ions, not a lower one.

Acetic acid (CH$_3$COOH) is a weak acid and does not typically dissociate 100%. Even if hypothetically 100% dissociated, both HCl and CH$_3$COOH are monoprotic acids producing equal concentrations of ions, resulting in similar conductivities rather than a 2:1 ratio.

This option reverses the required order. H$_2$SO$_4$ (diprotic) would have a higher conductivity than HNO$_3$ (monoprotic), meaning Solution Y would be more conductive than Solution X, contradicting the problem statement.

H$_3$PO$_4$ is a weak acid. If assumed to be 100% dissociated, it would release three H$^+$ ions, likely resulting in a conductivity ratio closer to 3:1 compared to the monoprotic HNO$_3$.

This incorrectly assumes the pH halves upon dilution. The solution is a buffer, so its pH depends on the ratio of acid to conjugate base, not just the total volume.

Diluting a buffer does not decrease its pH, as the ratio of the weak acid to its conjugate base remains unchanged.

This incorrectly assumes the pH increases upon dilution (like an unbuffered acid). Because this is a buffer solution, it resists changes to its pH when diluted.

Statement (iii) is incorrect. While it is true that $[H^+] = [OH^-]$ in pure water, water only ionises to a very small extent, not completely.

Statements (iii) and (iv) are both incorrect. Water does not ionise completely, and its self-ionisation produces equal amounts of hydrogen and hydroxide ions, not twice as many.

This option is incorrect because statements (iii) and (iv) are false. Water only partially ionises and produces a 1:1 ratio of hydrogen to hydroxide ions.

This order incorrectly ranks HCN as stronger than $CH_3COOH$. Although HCN has a higher concentration ($1.5 \times 10^{-1}$ M) than $CH_3COOH$ ($1.0 \times 10^{-1}$ M), it produces a much higher pH (5.0 vs 2.8), indicating it dissociates less and is the weaker acid.

This option lists the acids in reverse order, from weakest to strongest. A lower pH indicates a higher concentration of $H^+$ ions and thus a stronger acid.

This option incorrectly identifies the weakest acid (HCN) as the strongest. The data shows HCN has the highest pH (5.0) despite having the highest concentration, making it the weakest acid in the group.

Incorrect. A conjugate acid-base pair must consist of two species that differ by exactly one proton ($\text{H}^+$), which is not the case for these two molecules.

Incorrect. These are the two products of the forward reaction. They do not differ by a single proton, so they are not a conjugate acid-base pair.

Incorrect. These species do not differ by a single proton ($\text{H}^+$), so they cannot be a conjugate acid-base pair.

While ethanoic acid is indeed a weak acid, a concentration of 10.0 M is very high, making it a concentrated solution rather than a dilute one.

Ethanoic acid is a weak acid because it only partially dissociates in water, and 10.0 M is a high concentration, not a dilute one.

Although the solution is concentrated (10.0 M), ethanoic acid is classified as a weak acid, not a strong acid, because it does not completely ionize in solution.

Ammonia ($NH_3$) acts as a weak base because it accepts protons, rather than donating multiple protons like a polyprotic acid.

Ammonium phosphate is an ionic salt composed of ammonium and phosphate ions, not a polyprotic acid molecule.

Acetic acid ($CH_3COOH$) is a monoprotic acid because only the single hydrogen in the carboxyl group ($-COOH$) is ionizable.

This represents the hydrogen ion concentration ($[H^+] = 10^{-pH}$), not the hydroxide ion concentration.

This incorrectly uses a positive exponent for the pOH value. The concentration must be calculated as $10^{-pOH}$, not $10^{pOH}$.

This incorrectly uses a positive exponent for the pH value instead of calculating the hydroxide concentration using $10^{-pOH}$.