Answer
Hydroxide concentration : $[OH^-] = 4.571 \times 10^{-4}M$
- $4.895 \times 10^{-3}$g of $Ba(OH)_2$.
Work Step by Step
1. Use the pH to find the concentration of $[OH^-]$
pH + pOH = 14
10.66 + pOH = 14
pOH = 3.34
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{- 3.34}$
$[OH^-] = 4.571 \times 10^{- 4}$
2. $Ba(OH)_2$ is a strong base with 2 $OH^-$ in each molecule, therefore:
$[OH^-] = 2 * [Ba(OH)_2]$
$4.571 \times 10^{-4} = 2 * [Ba(OH)_2]$
$\frac{4.571 \times 10^{-4}}{2} = [Ba(OH)_2]$
$[Ba(OH)_2] = 2.286 \times 10^{-4}M$
3. Calculate the number of moles:
$n(moles) = concentration(M) * volume(L)$
$n(moles) = 2.286\times 10^{- 4} * 0.125$
$n(moles) = 2.857\times 10^{- 5}$
4. Find the mass value in grams:
Molar Mass ($Ba(OH)_2$):
137.3* 1 + 16* 2 + 1.01* 2 = 171.32g/mol
$mass(g) = mm(g/mol) * n(moles)$
$mass(g) = 171.32 * 2.857\times 10^{- 5}$
$mass(g) = 4.895\times 10^{- 3}$
