SATHEE: Equilibrium - Result Question 47

Equilibrium - Result Question 47

####49. The hydrogen ion concentration of a $10^{- 8} M$ $H C l$ aqueous solution at $298 K (K_{w} = 10^{- 14})$ is

[2006]

(a) $11 \times 10^{- 8} M$

(b) $9.525 \times 10^{- 8} M$

(d) $1.0 \times 10^{- 6} M$

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Solution:

(a) For a solution of $10^{- 8} M H C l, [H^{+}] = 10^{- 8}$

$[H^{+}]$ of water $= 10^{- 7}$

Total $[H^{+}] = 10^{- 7} + 10^{- 8} = 10 \times 10^{- 8} + 10^{- 8}$

$10^{- 8} (10 + 1) = 11 \times 10^{- 8}$

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Equilibrium - Result Question 47

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