91影视

(a)

The equilibrium reactions are

$\begin{array}{rcl}& & Ca{\left(OH\right)}_2\stackrel{K_{sp}}{鈬�}C{a}^{2+}+2O{H}^{-}---\left(1\right)\\ & & C{a}^{2+}+2O{H}^{-}\stackrel{K_1}{鈬�}CaO{H}^{+}----\left(2\right)\\ & & H_{2}O\stackrel{K_w}{鈬�}{H}^{+}+O{H}^{-}-----\left(3\right)\end{array}$

$\begin{array}{rcl}{K}_{sp}& =& 6.5脳{10}^{-6}\\ K_1& =& 2.0脳{10}^1\\ K_w& =& 1.0脳{10}^{-14}\\ K_{sp}& =& \left[C{a}^{2+}\right]{纬}_{C{a}^{2+}}{\left[O{H}^{-}\right]}^2{{纬}_{O{H}^{-}}}^2\\ K_1& =& \frac{\left[CaO{H}^{+}\right]}{\left[C{a}^{2+}\right]{纬}_{C{a}^{2+}}\left[O{H}^{-}\right]{纬}_{O{H}^{-}}}\\ K_w& =& \left[H^{+}\right]{纬}_{H^{+}}\left[O{H}^{-}\right]{纬}_{O{H}^{-}}\end{array}$

Charge balance is written as follows

$\left[2C{a}^{2+}\right]+\left[CaO{H}^{+}\right]+\left[H^{+}\right]=\left[O{H}^{-}\right]-----\left(4\right)$

Mass balance is written as follows

$2\left\{\left[2C{a}^{2+}\right]+\left[CaO{H}^{+}\right]\right\}+\left[H^{+}\right]=\left[O{H}^{-}\right]+\left[CaO{H}^{+}\right]----\left(5\right)$

(b)

Neglecting the activity coefficient and neglecting [H⁺] in the charge balance because [H⁺] <<[OH^-] in basic solution.

Therefore, the charge balance becomes

$\left[2C{a}^{2+}\right]+\left[CaO{H}^{+}\right]=\left[O{H}^{-}\right]----\left(6\right)$

After substituting [CaOH⁺] we get

$\begin{array}{rcl}\left[2C{a}^{2+}\right]+K_1\left[C{a}^{2+}\right]\left[O{H}^{-}\right]& =& \left[O{H}^{-}\right]\\ C{a}^{2+}& =& \frac{\left[O{H}^{-}\right]}{2+K_1\left[O{H}^{-}\right]}\\ K_{sp}& =& \left[C{a}^{2+}\right]{\left[O{H}^{-}\right]}^2\\ & =& \frac{{\left[O{H}^{-}\right]}^3}{2+K_1\left[O{H}^{-}\right]}\end{array}$

Equation 7 must be solved using spreadsheet and by guessing the[OH^-] value

Using SOLVER the following spreadsheet was obtained

The values obtained are as follows

[Ca²⁺] =0.01 M, [OH^-]=0.0253 M, [CaOH⁺]= 0.005 M, and [H⁺闭=3.94脳10^-13M

The fraction of hydrolysis =( [CaOH⁺]/{[Ca²⁺] + [CaOH⁺]})=0.333

Total dissolved Ca=0.01+0.005=0.015M

Formula mass of Ca(OH)₂ 74.09 g/mol

Solubility=74.09脳0.015 g/L= 1.11 g/L