## Chemistry and Chemical Reactivity (9th Edition)

Hydroxide concentration : $[OH^-] = 4.571 \times 10^{-4}M$ - $4.895 \times 10^{-3}$g of $Ba(OH)_2$.
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 $mass(g) = mm(g/mol) * n(moles)$ $mass(g) = 171.32 * 2.857\times 10^{- 5}$ $mass(g) = 4.895\times 10^{- 3}$