The number of hydroxyl ions in 10 text{ cm}^3 | vedclass

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(C) Given: $[H^{+}] = 0.2 	ext{ M}$,Volume $V = 10 	ext{ cm}^3 = 10^{-2} 	ext{ L}$.At $298 	ext{ K}$,the ionic product of water is $K_w = [H^{+}][

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$3 	imes 10^{12}$

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(C) Given: $[H^{+}] = 0.2 	ext{ M}$,Volume $V = 10 	ext{ cm}^3 = 10^{-2} 	ext{ L}$.At $298 	ext{ K}$,the ionic product of water is $K_w = [H^{+}][OH^{-}] = 10^{-14}$.$[OH^{-}] = \frac{K_w}{[H^{+}]} = \frac{10^{-14}}{0.2} = 5 	imes 10^{-14} 	ext{ mol/L}$.Number of $OH^{-}$ ions $= [OH^{-}] 	imes V 	imes N_A$.$= (5 	imes 10^{-14} 	ext{ mol/L}) 	imes (10^{-2} 	ext{ L}) 	imes (6.022 	imes 10^{23} 	ext{ ions/mol}) \approx 3.011 	imes 10^8$.Note: The calculated value is $3.011 	imes 10^8$. The provided option $(c)$ $3 	imes 10^{12}$ is mathematically inconsistent with the given parameters.

The number of hydroxyl ions in $10 \text{ cm}^3$ of $0.2 \text{ M}$ $HCl$ solution is

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