Assuming equal concentrations, rank these solutions by pH. Ca(OH)2(aq) KOH(aq) NH3(aq) HClO4(aq) HF(aq)?

Assume that the concentrations of all of the solutions are M mol/L.

Ca(OH)₂ is a strong alkali. 1 mole of Ca(OH)₂ dissociates to form 2 moles of OH⁻ ions.
Hence, [OH⁻] in M mol/L Ca(OH)₂ solution = 2M mol/L

KOH is a strong alkali. 1 mole of KOH dissociates to form 1 mole of OH⁻ ions.
Hence, [OH⁻] in M mol/L KOH solution = M mol/L

NH₃ is a weak alkali, which partially dissociates to form OH⁻ ions.
Hence, [OH⁻] in M mol/L NH₃ solution < M mol/L

In alkalis, the higher the concentration of OH⁻ ions, the higher the pH is.
The concentration of OH⁻ ions : Ca(OH)₂ > KOH > NH₃
pH of the solution : Ca(OH)₂ > KOH > NH₃

HClO₄ is a strong acid. 1 mole of HClO₄ dissociates to form 1 mole of H⁺ ions.
Hence, [H⁺] in M mol/L HClO₄ solution = M mol/L

HF is a weak acid, which partially dissociation to form H⁺ ios.
Hence, [H⁺] in M mol/L HF < M mol/L

In acids, the higher the concentration of H⁺ ions, the lower the pH is.
The concentration of H⁺ : HF < HClO₄
pH of the solution : HF > HClO₄

pH of alkalis is higher than that of acids.
Conclusively, pH of the solution : Ca(OH)₂ > KOH > NH₃ > HF > HClO₄

KOH(aq)
Ca(OH)2(aq)
NH3(aq)
HF(aq)
HClO4(aq)