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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 \((log\,2 = 0.3)\)
- A \(14\)
- B \(13.70\)
- C \(13\)
- D \(12.70\)
Answer & Solution
Correct Answer
(D) \(12.70\)
Step-by-step Solution
Detailed explanation
Given \([O{H^ - }] = 5 \times {10^{ - 2}}\) \(\therefore \,pOH = - \log \,5 \times {10^{ - 2}}\) \( = - \log \,5 + 2\log \,10 = 1.30\) \(\because \,pH + pOH = 14\) \(\because \,pH = 14 - pOH\) \( = 14 - 1.30 = 12.70\)
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