Answer
$[OH^-] = 4.571 \times 10^{- 4}M$
0.008817g of $Ba(OH)_2$.
Work Step by Step
1. Calculate [OH^-]:
pH + pOH = 14
10.66 + pOH = 14
pOH = 3.34
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{- 3.34}$
$[OH^-] = 4.571 \times 10^{- 4}$
2. Since $Ba(OH)_2$ is a strong base with 2 OH in each molecule: $[OH^-] = 2 * [Ba(OH)_2]$
$ 4.571\times 10^{- 4} = 2 * [Ba(OH)_2]$
$ \frac{ 4.571\times 10^{- 4}}{ 2} = [Ba(OH)_2]$
$ 2.285\times 10^{- 4}M = [Ba(OH)_2]$
3. Calculate the number of moles: $(Ba(OH)_2)$
$n(moles) = concentration(M) * volume(L)$
- 250ml = 0.250L
$n(moles) = 2.285\times 10^{- 4} * 0.25$
$n(moles) = 5.714\times 10^{- 5}$
4. Find the mass value in grams: $(Ba(OH)_2)$
137.3* 1 + 2 * ( 16* 1 + 1.01* 1 ) = 154.31g/mol
$mass(g) = mm(g/mol) * n(moles)$
$mass(g) = 154.31 * 5.714\times 10^{- 5}$
$mass(g) = 8.817\times 10^{- 3}$