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JIPMER 2005
Exam
JIPMER
Year
2005
Subject
Chemistry
Chapter
Ionic Equilibrium

JIPMER 2005 · Chemistry · Ionic Equilibrium

The pH of a 10^ -10 M-NaOH solution is

The pH of a 10−10M−NaOH10^{-10} \mathrm{M}-\mathrm{NaOH} solution is

Options

  1. A10
  2. B7.01
  3. C6.99
  4. D4

Correct answer

B. 7.01

Step-by-step solution

When the solution is very dilute, the concentration of OH−\mathrm{OH}^{-} produced from water cannot be neglected. Hence, [OH−]=10−10+10−7 (obtained from water) =10−7(0.001+1)=1.001×10−7∴pOH=−log⁡[OH−]=−log⁡(1.001×10−7)=7−0.01=6.99∴pH=14−pOH=14−6.99=7.01\begin{aligned} & {\left[\mathrm{OH}^{-}\right]=10^{-10}+10^{-7}} \\ & \text{ (obtained from water) } \\ & =10^{-7}(0.001+1)=1.001 \times 10^{-7} \\ & \therefore \quad \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right] \\ & =-\log \left(1.001 \times 10^{-7}\right) \\ & =7-0.01=6.99 \\ & \therefore \quad \mathrm{pH}=14-\mathrm{pOH} \\ & =14-6.99=7.01 \\ & \end{aligned}
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