Answer
$[OH^-] = 1.62 \times 10^{-2}M$
$[H_3O^+] = 6.17 \times 10^{- 13}M$
$[Ca^{2+}] = 8.10 \times 10^{-3}M$
Work Step by Step
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}$
$[H_3O^+] = \frac{10^{-14}}{ 1.62 \times 10^{- 2}}$
$[H_3O^+] = 6.173 \times 10^{- 13}$
