JIPMER 2004
- Exam
- JIPMER
- Year
- 2004
- Subject
- Chemistry
- Chapter
- Ionic Equilibrium
JIPMER 2004 · Chemistry · Ionic Equilibrium
The pH of 10^ -8 M NaOH is
The pH of
10−8MNaOH is
Options
- A8.0
- B6.0
- C6.96
- D7.04
Correct answer
D. 7.04
Step-by-step solution
For
10−8MNaOH,
[OH−]=10−8+10−7 (from water)
=11×10−8∴pOH=−log[OH−]=−log(11×10−8)=8−1.04=6.96∴pH=14−6.96=7.04
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