Question

Which of the following aqueous solutions has highest pH? (Given \(\log 2 = 0.30\), \(\log 3 = 0.48\), \(\log 4 = 0.60\), \(\log 5 = 0.70\))

• A. 0.2 M \({Ba(OH)}_2\)
• B. 0.02 N \({Ba(OH)}_2\)
• C. 0.1 M NaOH
• D. 0.05 M \({Ba(OH)}_2\)

Hint:

In strong bases like \({Ba(OH)}_2\), OH\(^-\) concentration is twice the molarity because of two hydroxide ions per formula unit.

Answer 1

The Correct Option is A

Calculate pOH for each: 
- For \(0.2\, M\, {Ba(OH)}_2\), \([{OH}^-] = 2 \times 0.2 = 0.4\, M\) \[ pOH = -\log(0.4) = -\log(4 \times 10^{-1}) = -( \log 4 + \log 10^{-1} ) = -(0.60 - 1) = 0.4 \] \[ pH = 14 - pOH = 14 - 0.4 = 13.6 \] - For \(0.02\, N\, {Ba(OH)}_2\), \([{OH}^-] = 2 \times 0.02 = 0.04\), \[ pOH = -\log(0.04) = -\log(4 \times 10^{-2}) = -(0.60 - 2) = 1.4 \] \[ pH = 12.6 \] - For \(0.1\, M\, {NaOH}\), \([{OH}^-] = 0.1\), \[ pOH = 1 \Rightarrow pH = 13 \] - For \(0.05\, M\, {Ba(OH)}_2\), \([{OH}^-] = 2 \times 0.05 = 0.1\), \[ pOH = 1 \Rightarrow pH = 13 \] Highest pH is 13.6 for 0.2 M \({Ba(OH)}_2\).