At 298 , K,what is the concentration of hydro | vedclass

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(B) For a very dilute solution of a strong acid,the contribution of $H^+$ ions from the auto-ionization of water cannot be neglected.Let the total con

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

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(B) For a very dilute solution of a strong acid,the contribution of $H^+$ ions from the auto-ionization of water cannot be neglected.Let the total concentration of $[H^+] = x$.The source of $H^+$ ions is from $HCl$ $(10^{-8} \, M)$ and water $(x - 10^{-8} \, M)$.The concentration of $OH^-$ ions from water is also $(x - 10^{-8} \, M)$.Using the ionic product of water: $K_W = [H^+][OH^-] = 10^{-14}$.$x(x - 10^{-8}) = 10^{-14}$.$x^2 - 10^{-8}x - 10^{-14} = 0$.Using the quadratic formula $x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{10^{-8} + \sqrt{(10^{-8})^2 - 4(1)(-10^{-14})}}{2}$.$x = \frac{10^{-8} + \sqrt{10^{-16} + 4 	imes 10^{-14}}}{2}$.$x = \frac{10^{-8} + \sqrt{10^{-16} + 400 	imes 10^{-16}}}{2} = \frac{10^{-8} + \sqrt{401 	imes 10^{-16}}}{2}$.$x = \frac{10^{-8} + 20.025 	imes 10^{-8}}{2} = \frac{21.025 	imes 10^{-8}}{2} = 1.05125 	imes 10^{-7} \, M$.Rounding to the nearest option,the correct answer is $1.0525 	imes 10^{-7} \, M$.

At $298 \, K$,what is the concentration of hydrogen ions in a $10^{-8} \, M \, HCl$ aqueous solution? $(K_W = 10^{-14})$

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