0.023 g of sodium metal is reacted with 100 | vedclass

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(D) The chemical reaction is: $2Na + 2H_2O ightarrow 2NaOH + H_2$Number of moles of $Na = \frac{0.023 \ g}{23 \ g/mol} = 1 	imes 10^{-3} \ mol$.Acc

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$12$

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(D) The chemical reaction is: $2Na + 2H_2O ightarrow 2NaOH + H_2$Number of moles of $Na = \frac{0.023 \ g}{23 \ g/mol} = 1 	imes 10^{-3} \ mol$.According to the stoichiometry,$2 \ mol$ of $Na$ produces $2 \ mol$ of $NaOH$.Therefore,$1 	imes 10^{-3} \ mol$ of $Na$ produces $1 	imes 10^{-3} \ mol$ of $NaOH$.The volume of the solution is $100 \ cm^3 = 0.1 \ L$.Concentration of $[OH^-] = \frac{1 	imes 10^{-3} \ mol}{0.1 \ L} = 1 	imes 10^{-2} \ M$.$pOH = -\log[OH^-] = -\log(1 	imes 10^{-2}) = 2$.Since $pH + pOH = 14$,$pH = 14 - 2 = 12$.

$0.023 \ g$ of sodium metal is reacted with $100 \ cm^{3}$ of water. The $pH$ of the resulting solution is

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