How many H^{+} ions are present in 2 mL of a | vedclass

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(B) Given $pH = 13$. Since $pH = -\log[H^{+}]$,we have $[H^{+}] = 10^{-pH} = 10^{-13} \ M$. Volume of solution $= 2 \ mL = 2 	imes 10^{-3} \ L$. Mole

Vedclass · Accepted Answer

$12.04 	imes 10^7$

Vedclass · Answer

(B) Given $pH = 13$. Since $pH = -\log[H^{+}]$,we have $[H^{+}] = 10^{-pH} = 10^{-13} \ M$. Volume of solution $= 2 \ mL = 2 	imes 10^{-3} \ L$. Moles of $H^{+}$ ions $= 	ext{Molarity} 	imes 	ext{Volume in Litres} = 10^{-13} 	imes 2 	imes 10^{-3} = 2 	imes 10^{-16} \ mol$. Number of $H^{+}$ ions $= 	ext{Moles} 	imes N_A = 2 	imes 10^{-16} 	imes 6.022 	imes 10^{23} = 12.044 	imes 10^7$ ions.

How many $H^{+}$ ions are present in $2 \ mL$ of a solution whose $pH$ is $13$?

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