The moles of H^{+} from H_2O alone in a 1 L, | vedclass

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(C) Let the concentration of $HCl$ be $C = \sqrt{5} 	imes 10^{-7} \ M = 2.23 	imes 10^{-7} \ M$. Since $HCl$ is a strong acid,it dissociates complet

Vedclass · Accepted Answer

$3.85 	imes 10^{-8}$

Vedclass · Answer

(C) Let the concentration of $HCl$ be $C = \sqrt{5} 	imes 10^{-7} \ M = 2.23 	imes 10^{-7} \ M$. Since $HCl$ is a strong acid,it dissociates completely: $[H^{+}]_{HCl} = C = 2.23 	imes 10^{-7} \ M$. Let $x$ be the concentration of $H^{+}$ ions produced by the dissociation of $H_2O$. The total concentration of $H^{+}$ ions is $[H^{+}]_{total} = [H^{+}]_{HCl} + [H^{+}]_{H_2O} = C + x$. The concentration of $OH^{-}$ ions is $[OH^{-}] = x$ (since $H_2O$ produces equal moles of $H^{+}$ and $OH^{-}$). At $25 \ ^\circ C$,the ionic product of water is $K_w = [H^{+}][OH^{-}] = 10^{-14}$. Substituting the values: $(C + x)(x) = 10^{-14}$. $x^2 + Cx - 10^{-14} = 0$. Using the quadratic formula $x = \frac{-C + \sqrt{C^2 + 4K_w}}{2}$: $x = \frac{-2.23 	imes 10^{-7} + \sqrt{(2.23 	imes 10^{-7})^2 + 4 	imes 10^{-14}}}{2}$. $x = \frac{-2.23 	imes 10^{-7} + \sqrt{4.97 	imes 10^{-14} + 4 	imes 10^{-14}}}{2} = \frac{-2.23 	imes 10^{-7} + \sqrt{8.97 	imes 10^{-14}}}{2}$. $x = \frac{-2.23 	imes 10^{-7} + 2.995 	imes 10^{-7}}{2} = \frac{0.765 	imes 10^{-7}}{2} = 0.3825 	imes 10^{-7} = 3.825 	imes 10^{-8} \ M$. Since the volume is $1 \ L$,the moles of $H^{+}$ from $H_2O$ is $3.825 	imes 10^{-8} \ mol$,which is approximately $3.85 	imes 10^{-8} \ mol$.

The moles of $H^{+}$ from $H_2O$ alone in a $1 \ L$,$\sqrt{5} \times 10^{-7} \ M$ $HCl$ solution at $25 \ ^\circ C$ is ( $\sqrt{5} = 2.23$ )

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