SATHEE: Equilibrium Question 308

Equilibrium Question 308

Question: The concentration of $[H^{+}]$ and concentration of $[O H^{-}]$ of a 0.1 aqueous solution of 2% ionised weak acid is [Ionic product of water $= 1 \times 10^{- 14}]$ [DPMT 2004; CBSE PMT 1999]

Options:

A) $2 \times 10^{- 3}$ M and $5 \times 10^{- 12}$ M

B) $1 \times 10^{3} M a n d 3 \times 10^{- 11} M$

C) $0.02 \times 10^{- 3} M a n d 5 \times 10^{- 11} M$

D) $3 \times 10^{- 2} M a n d 4 \times 10^{- 13} M$

Show Answer

Answer:

Correct Answer: A

Solution:

Given that Concentration of solution =.1 Degree of ionisation $= 2$ Ionic product of water $= 1 \times 10^{- 14}$ Concentration of $[H^{+}]$ = Concentration of solution X degree of ionisation $= .1 \times .02 = 2 \times 10^{- 3} M$ Concentration of $[O H^{-}] = \frac{Ionic product of water}{[H^{+}]}$

$= \frac{1 \times 10^{- 14}}{2 \times 10^{- 3}} = 0.5 \times 10^{- 11} = 5 \times 10^{- 12} M$ .

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Equilibrium Question 308

Question: The concentration of [H+] and concentration of [OH−] of a 0.1 aqueous solution of 2% ionised weak acid is [Ionic product of water =1×10−14] [DPMT 2004; CBSE PMT 1999]

Options:

Answer:

Solution:

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Question: The concentration of $[H^{+}]$ and concentration of $[O H^{-}]$ of a 0.1 aqueous solution of 2% ionised weak acid is [Ionic product of water $= 1 \times 10^{- 14}]$ [DPMT 2004; CBSE PMT 1999]