## General Chemistry (4th Edition)

$[OH^-] = 1.62 \times 10^{-2}M$ $[H_3O^+] = 6.17 \times 10^{- 13}M$ $[Ca^{2+}] = 8.10 \times 10^{-3}M$
1. Calculate the molar mass: 40.08* 1 + 2 * ( 16* 1 + 1.01* 1 ) = 74.10g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.600}{ 74.10}$ $n(moles) = 8.097 \times 10^{-3}$ 3. Find the concentration in mol/L: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 8.097 \times 10^{-3}}{ 1}$ $C(mol/L) = 8.097 \times 10^{-3}$ - This is the $Ca(OH)_2$ concentration. Since each $Ca(OH)_2$ has 2 $OH^-$: $[OH^-] = [Ca(OH)_2] * 2 = 8.097 \times 10^{-3} * 2 = 1.62 \times 10^{-2}M$ Since each $Ca(OH)_2$ has 1 $Ca^{2+}:$ $[Ca^{2+}] = [Ca(OH)_2] = 8.097 \times 10^{-3}$ - Now, use the $K_w$ to find the $[H_3O^+]$ value: $[OH^-] * [H_3O^+] = Kw = 10^{-14}$ $1.62 \times 10^{- 2} * [H_3O^+] = 10^{-14}$ $[H_3O^+] = \frac{10^{-14}}{ 1.62 \times 10^{- 2}}$ $[H_3O^+] = 6.173 \times 10^{- 13}$