SATHEE: Equilibrium Question 424

Equilibrium Question 424

Question: The $p H$ of $0.0001 N$ solution of $K O H$ will be [BHU 1997; CET Pune 1998]

Options:

A) 4

B) 6

C) 10

D) 12

Show Answer

Answer:

Correct Answer: C

Solution:

$[O H^{-}] = 0.0001 N,$

$p O H = 4$ , $p H + p O H = 14$

$p H = 14 - p O H$

$= 14 - 4 = 10$

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Equilibrium Question 424

Question: The pH of 0.0001N solution of KOH will be [BHU 1997; CET Pune 1998]

Options:

Answer:

Solution:

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Question: The $p H$ of $0.0001 N$ solution of $K O H$ will be [BHU 1997; CET Pune 1998]