Equilibrium - Result Question 47
####49. The hydrogen ion concentration of a $10^{-8} M$ $HCl$ aqueous solution at $298 K(K_w=10^{-14})$ is
[2006]
(a) $11 \times 10^{-8} M$
(b) $9.525 \times 10^{-8} M$
(c) $1.0 \times 10^{-8} M$
(d) $1.0 \times 10^{-6} M$
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Solution:
- (a) For a solution of $10^{-8} M HCl,[H^{+}]=10^{-8}$
$[H^{+}]$of water $=10^{-7}$
Total $[H^{+}]=10^{-7}+10^{-8}=10 \times 10^{-8}+10^{-8}$
$10^{-8}(10+1)=11 \times 10^{-8}$
