The number of OH^{-} ions in 1 mL of solutio | vedclass

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(C) Given $pH = 4$,so $pOH = 14 - 4 = 10$.The concentration of $OH^{-}$ ions is $[OH^{-}] = 10^{-pOH} = 10^{-10} \ M$.Since $1 \ M = 1 \ mol/L = 1 \ m

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$6.02 	imes 10^{10}$

Vedclass · Answer

(C) Given $pH = 4$,so $pOH = 14 - 4 = 10$.The concentration of $OH^{-}$ ions is $[OH^{-}] = 10^{-pOH} = 10^{-10} \ M$.Since $1 \ M = 1 \ mol/L = 1 \ mol/1000 \ mL$,the number of moles of $OH^{-}$ in $1 \ mL$ is $\frac{10^{-10}}{1000} = 10^{-13} \ mol$.The number of $OH^{-}$ ions is calculated by multiplying the number of moles by Avogadro's number $(N_A = 6.023 	imes 10^{23} \ mol^{-1})$:$	ext{Number of ions} = 10^{-13} 	imes 6.023 	imes 10^{23} = 6.023 	imes 10^{10}$.

The number of $OH^{-}$ ions in $1 \ mL$ of solution having $pH = 4$ is :-

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