ExamBro
ExamBro
JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)

\(NaOH\) is a strong base. What will be \(pH\) of \(5.0 \times 10^{-2}\,M\) \(NaOH\) solution \((log\,2 = 0.3)\)

  1. A \(14\)
  2. B \(13.70\)
  3. C \(13\)
  4. D \(12.70\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(12.70\)

Step-by-step Solution

Detailed explanation

Given \([O{H^ - }] = 5 \times {10^{ - 2}}\) \(\therefore \,pOH = - \log \,5 \times {10^{ - 2}}\) \( = - \log \,5 + 2\log \,10 = 1.30\) \(\because \,pH + pOH = 14\) \(\because \,pH = 14 - pOH\) \( = 14 - 1.30 = 12.70\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app