The hydrogen ion concentration of a 10^{-8} | vedclass

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Question

(C) In an aqueous solution of $10^{-8} \ M \ HCl$,the total $[H^{+}]$ is the sum of $[H^{+}]$ from $HCl$ and $[H^{+}]$ from the dissociation of water.

Vedclass · Accepted Answer

$1.0525 	imes 10^{-7} \ M$

Vedclass · Answer

(C) In an aqueous solution of $10^{-8} \ M \ HCl$,the total $[H^{+}]$ is the sum of $[H^{+}]$ from $HCl$ and $[H^{+}]$ from the dissociation of water.Let $[H^{+}] = x$.From $HCl$,$[H^{+}] = 10^{-8} \ M$.From water,$[H^{+}] = [OH^{-}] = x - 10^{-8} \ M$.Using the ionic product of water,$K_w = [H^{+}][OH^{-}] = 10^{-14}$.$x(x - 10^{-8}) = 10^{-14}$$x^{2} - 10^{-8}x - 10^{-14} = 0$Solving this quadratic equation using the quadratic formula $x = \frac{-b + \sqrt{b^{2} - 4ac}}{2a}$:$x = \frac{10^{-8} + \sqrt{(10^{-8})^{2} - 4(1)(-10^{-14})}}{2}$$x = \frac{10^{-8} + \sqrt{10^{-16} + 4 	imes 10^{-14}}}{2}$$x = \frac{10^{-8} + \sqrt{10^{-16} + 400 	imes 10^{-16}}}{2}$$x = \frac{10^{-8} + \sqrt{401 	imes 10^{-16}}}{2}$$x = \frac{10^{-8} + 20.025 	imes 10^{-8}}{2}$$x = \frac{21.025 	imes 10^{-8}}{2} = 1.05125 	imes 10^{-7} \ M \approx 1.0525 	imes 10^{-7} \ M$.

The hydrogen ion concentration of a $10^{-8} \ M \ HCl$ aqueous solution at $298 \ K$ $(K_w = 10^{-14})$ is

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