Answer
$ 2.18 \times 10^{-3}$ grams.
Work Step by Step
1. Find the concentration of $[OH^-]$:
$pOH + pH = 14$
$pOH + 10 = 14$
$pOH = 4$
$[OH^-] = 10^{-pOH} = 10^{-4}$
2. Since $NaOH$ is a strong base:
$[NaOH] = [OH^-] = 10^{-4}$
3. With the concentration and the volume, find the number of moles:
$546 ml = 0.546L$
$n(moles) = concentration(M) \times volume(L)$
$n(moles) = 10^{-4} \times 0.546 = 5.46 \times 10^{-5}$
4. Find the molar mass of $NaOH$.
Na: 23
O: 16
H: 1
NaOH = 40 g/mol.
5. Find the mass of $5.46 \times 10^{-5}$ moles of $NaOH$.
$mass(g) = n(moles) \times molar mass$
$mass(g) = 5.46 \times 10^{-5} \times 40 = 2.18 \times 10^{-3}g$
