SATHEE: Equilibrium Question 425

Equilibrium Question 425

Question: Given that the dissociation constant for $H_{2} O$ is $K_{w} = 1 \times 10^{- 14}$

$m o l e^{2} l i t r e^{- 2},$ what is the $H C l$ of a $0.001 m o l a r K O H$ solution [MP PET 1995; MP PET/PMT 1998]

Options:

A) $10^{- 11}$

B) 3

C) 14

D) 11

Show Answer

Answer:

Correct Answer: D

Solution:

0.001 M KOH solution $[O H^{-}] = 0.001, M = 1 \times 10^{- 3} M$

$[H^{+}] \times [O H^{-}] = 1 \times 10^{- 14}$

$[H^{+}] = \frac{1 \times 10^{- 14}}{[O H^{-}]}$

$[H^{+}] = \frac{1 \times 10^{- 14}}{1 \times 10^{- 3}}$

$= 1 \times 10^{- 14} \times 10^{+ 3}$

$[H^{+}] = 10^{- 11}$ M $p H = 11$

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Equilibrium Question 425

Question: Given that the dissociation constant for H2O is Kw=1×10−14

Options:

Answer:

Solution:

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Question: Given that the dissociation constant for $H_{2} O$ is $K_{w} = 1 \times 10^{- 14}$