First slide

Ionic equilibrium

Question

$pH of 10^{- 8} M NaOH is$

Easy

A

8.0
B

6.0
C

6.98
D

7.02

Solution

$[NaOH] = 10^{- 8} M [{OH}^{-}] = 10^{- 8} M [H^{+}] = 10^{- 6} M pH \neq 6 since, solution is basic [{OH}^{-}] of water in this dilute solution should be added ∴ {[{OH}^{-}]}_{total} = \underset{H_{2} O}{\underset{⏟}{1 \times 10^{- 7} M}} + \underset{NaOH}{\underset{⏟}{1 \times 10^{- 8} M}} = 11 \times 10^{- 8} ∴ pOH = 8 - \log 11 pH = 14 - pOH = 6 + \log 11 = 7 .02$

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