The number of H^{+} ions present in 10 mL of | vedclass

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(D) Given: $pH = 12$ and Volume $V = 10 \ mL = 0.01 \ L$. Since $pH = -\log[H^{+}]$,we have $[H^{+}] = 10^{-pH} = 10^{-12} \ mol/L$. The number of mol

Vedclass · Accepted Answer

$6.023 	imes 10^{9}$

Vedclass · Answer

(D) Given: $pH = 12$ and Volume $V = 10 \ mL = 0.01 \ L$. Since $pH = -\log[H^{+}]$,we have $[H^{+}] = 10^{-pH} = 10^{-12} \ mol/L$. The number of moles of $H^{+}$ ions = $[H^{+}] 	imes V = 10^{-12} \ mol/L 	imes 0.01 \ L = 10^{-14} \ mol$. The number of $H^{+}$ ions = $	ext{moles} 	imes N_A = 10^{-14} 	imes 6.023 	imes 10^{23} = 6.023 	imes 10^{9}$.

The number of $H^{+}$ ions present in $10 \ mL$ of a solution with $pH = 12$ is ........

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