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JIPMER 2011
Exam
JIPMER
Year
2011
Subject
Chemistry
Chapter
Ionic Equilibrium

JIPMER 2011 · Chemistry · Ionic Equilibrium

10^ -6 M~ NaOH is diluted 100 times. The pH of the

10−6M NaOH10^{-6} \mathrm{M~} \mathrm{NaOH} is diluted 100 times. The pH of the

Options

  1. Abetween 7 and 8
  2. Bbetween 5 and 6
  3. Cbetween 6 and 7
  4. Dbetween 10 and 11

Correct answer

A. between 7 and 8

Step-by-step solution

[OH−]\left[\mathrm{OH}^{-}\right] in the diluted base =10−6102=10−8=\frac{10^{-6}}{10^2}=10^{-8} Total [OH−]=10−8+[OH−]of water =(10−8+10−7)M=10−8[1+10]M=11×10−8MpOH=−log⁡(11×10−8)=−log⁡11+8log⁡10=6.9586pH=14−6.9586=7.0414\begin{aligned} \text{ Total }\left[\mathrm{OH}^{-}\right] & =10^{-8}+\left[\mathrm{OH}^{-}\right] \text{of water } \\ & =\left(10^{-8}+10^{-7}\right) \mathrm{M} \\ & =10^{-8}[1+10] \mathrm{M} \\ & =11 \times 10^{-8} \mathrm{M} \\ \mathrm{pOH} & =-\log \left(11 \times 10^{-8}\right) \\ & =-\log 11+8 \log 10 \\ & =6.9586 \\ \mathrm{pH} & =14-6.9586 \\ & =7.0414\end{aligned}$$
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