How many grams of $NaOH$ are required to prepare $1 \ L$ of $NaOH$ solution with a $pH$ of $10.06$?
(N/A) $1$. The $pH$ of the solution is $10.06$. Since $NaOH$ is a strong base,we first calculate the $pOH$ using the relation: $pH + pOH = 14$.
$2$. $pOH = 14 - 10.06 = 3.94$.
$3$. The concentration of hydroxide ions $[OH^-]$ is given by $[OH^-] = 10^{-pOH} = 10^{-3.94} \approx 1.148 \times 10^{-4} \ M$.
$4$. Since $NaOH$ dissociates completely as $NaOH \rightarrow Na^+ + OH^-$,the molarity of $NaOH$ is equal to $[OH^-]$,which is $1.148 \times 10^{-4} \ mol/L$.
$5$. The molar mass of $NaOH$ is $23 + 16 + 1 = 40 \ g/mol$.
$6$. Mass of $NaOH = \text{Molarity} \times \text{Volume} \times \text{Molar Mass} = 1.148 \times 10^{-4} \ mol/L \times 1 \ L \times 40 \ g/mol = 4.592 \times 10^{-3} \ g$.
$2$. $pOH = 14 - 10.06 = 3.94$.
$3$. The concentration of hydroxide ions $[OH^-]$ is given by $[OH^-] = 10^{-pOH} = 10^{-3.94} \approx 1.148 \times 10^{-4} \ M$.
$4$. Since $NaOH$ dissociates completely as $NaOH \rightarrow Na^+ + OH^-$,the molarity of $NaOH$ is equal to $[OH^-]$,which is $1.148 \times 10^{-4} \ mol/L$.
$5$. The molar mass of $NaOH$ is $23 + 16 + 1 = 40 \ g/mol$.
$6$. Mass of $NaOH = \text{Molarity} \times \text{Volume} \times \text{Molar Mass} = 1.148 \times 10^{-4} \ mol/L \times 1 \ L \times 40 \ g/mol = 4.592 \times 10^{-3} \ g$.
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