0.023 g of sodium metal is reacted with 100 | vedclass

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Question

(D) The chemical reaction is: $2Na + 2H_{2}O \longrightarrow 2NaOH + H_{2}$ Molar mass of $Na = 23 \ g/mol$. Moles of $Na = \frac{0.023 \ g}{23 \ g/mo

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) The chemical reaction is: $2Na + 2H_{2}O \longrightarrow 2NaOH + H_{2}$ Molar mass of $Na = 23 \ g/mol$. Moles of $Na = \frac{0.023 \ g}{23 \ g/mol} = 0.001 \ mol$. According to the stoichiometry,$2 \ mol$ of $Na$ produces $2 \ mol$ of $NaOH$. Therefore,$0.001 \ mol$ of $Na$ produces $0.001 \ mol$ of $NaOH$. The volume of the solution is $100 \ cm^{3} = 0.1 \ L$. Concentration of $[OH^{-}] = \frac{0.001 \ mol}{0.1 \ L} = 0.01 \ M = 10^{-2} \ M$. $pOH = -\log[OH^{-}] = -\log(10^{-2}) = 2$. Since $pH + pOH = 14$,then $pH = 14 - 2 = 12$.

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

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