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JEE Main 201322 AprilOnline
Exam
JEE Main
Year
2013
Date
22 April
Mode
Online
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2013 · 22 April · Online · Chemistry · Ionic Equilibrium

NaOH is a strong base. What will be pH of 5.0 × 10^ -2 M NaOH...

NaOH\mathrm{NaOH} is a strong base. What will be pH\mathrm{pH} of 5.0×10−2MNaOH5.0 \times 10^{-2} \mathrm{M} \mathrm{NaOH} solution ? (log⁡2=0.3)(\log 2=0.3)

Options

  1. A14.0014.00
  2. B13.7013.70
  3. C13.0013.00
  4. D12.7012.70

Correct answer

D. 12.7012.70

Step-by-step solution

Given [OH−]=5×10−2\left[\mathrm{OH}^{-}\right]=5 \times 10^{-2}∴pOH=−log⁡5×10−2=−log⁡5+2log⁡10=1.30∵pH+pOH=14∵pH=14−pOH=14−1.30=12.70\begin{aligned}& \therefore \mathrm{pOH}=-\log5 \times10^{-2} \\ & \quad \quad=-\log 5+2 \log 10=1.30 \\ & \because \mathrm{pH}+\mathrm{pOH}=14 \\ & \because \mathrm{pH}=14-\mathrm{pOH}\\ & \quad=14-1.30=12.70 \end{aligned}$
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