If Solution X has a pH of 3 and Solution Y has a pH of 6, we can conclude that A - VCE - SSCE Chemistry - Question 4 - 2014 - Paper 1

From the previous calculations, we found: $[H^+]_{X} = 10^{-3}$ and $[H^+]_{Y} = 10^{-6}$. This implies: $\frac{[H^+]_{X}}{[H^+]_{Y}} = 1000$, indicating that Solution X contains much more [H+] than Solution Y, making this statement false.

Using the relationship between pH and pOH: $pOH = 14 - pH$,

For Solution X: $pOH_{X} = 14 - 3 = 11$, $[OH^-]_{X} = 10^{-11}$.

For Solution Y: $pOH_{Y} = 14 - 6 = 8$, $[OH^-]_{Y} = 10^{-8}$.

Now, we calculate: $\frac{[OH^-]_{Y}}{[OH^-]_{X}} = \frac{10^{-8}}{10^{-11}} = 10^{3} = 1000$,

Thus, [OH-] in Solution Y is not twice but 1000 times that of [OH-] in Solution X, making this statement false.

Stronger acids dissociate to give higher concentrations of [H+]. Since we have found that [H+] in Solution X (1000 times more than in Solution Y), Solution Y must not have a stronger acid than Solution X, thus this statement is also false.