Question

Determine the maximum mass fraction of calcium bicarbonate \(\left.\left[\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\right)\right]\) in water at \(350 \mathrm{~K}\).

Short answer

The approximate maximum mass fraction of calcium bicarbonate in water at 350 K is 1.12 x 10^{-3}.

Step-by-step solution

Find the solubility product constant, \(\mathrm{K}_{sp}\), for calcium bicarbonate.

Unfortunately, the \(\mathrm{K}_{sp}\) value for calcium bicarbonate at \(350 \mathrm{~K}\) is not readily available in standard reference tables. Therefore, we will assume that the \(\mathrm{K}_{sp}\) remains constant with temperature, using a known value at room temperature, \(25^\circ\mathrm{C}\) or \(298\mathrm{~K}\): $$ \mathrm{K}_{sp} = 2.7 \times 10^{-10} $$ Keep in mind that this assumption will lead to an approximate answer.

Write the balanced dissolution reaction of calcium bicarbonate in water.

The dissolution reaction of calcium bicarbonate in water is given by: $$ \mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2\mathrm{(s)}} \rightleftharpoons \mathrm{Ca}^{2+}_\mathrm{(aq)} + 2\mathrm{HCO}_{3\mathrm{^-}}_\mathrm{(aq)} $$

Determine the solubility of \(\left[\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\right]\) in water using \(\mathrm{K}_{sp}\) equation.

The \(\mathrm{K}_{sp}\) expression for this reaction is: $$ \mathrm{K}_{sp} = \left[\mathrm{Ca}^{2+}\right] \cdot \left[\mathrm{HCO}_{3\mathrm{^-}}\right]^2 $$ Let the solubility of \(\left[\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\right]\) in water be \(x \mathrm{~mol~L^{-1}}\). At equilibrium, we'll have \(\left[\mathrm{Ca}^{2+}\right] = x \mathrm{~mol~L^{-1}}\) and \(\left[\mathrm{HCO}_{3\mathrm{^-}}\right] = 2x \mathrm{~mol~L^{-1}}\). Plugging these into the \(\mathrm{K}_{sp}\) equation, we get: $$ 2.7 \times 10^{-10} = x \cdot (2x)^2 $$

Calculate the solubility of calcium bicarbonate in water.

To find the solubility of calcium bicarbonate, we need to solve the equation for \(x\): $$ 2.7 \times 10^{-10} = x \cdot (4x^2) \\ 2.7 \times 10^{-10} = 4x^3 \\ x^3 = \frac{2.7 \times 10^{-10}}{4} \\ x = \sqrt[3]{\frac{2.7 \times 10^{-10}}{4}} \\ x \approx 6.93 \times 10^{-4}\ \mathrm{mol~L^{-1}} $$ The solubility of calcium bicarbonate in water at \(350\ \mathrm{K}\) (approximated) is about \(6.93 \times 10^{-4}\ \mathrm{mol~L^{-1}}\).

Calculate the mass fraction of calcium bicarbonate in the solution.

The mass fraction of calcium bicarbonate in the solution can be calculated using the molar mass of calcium bicarbonate, \(162.11\ \mathrm{g~mol^{-1}}\) and the molar mass of water, \(18.015\ \mathrm{g~mol^{-1}}\). The mass of dissolved calcium bicarbonate per liter of water: $$ 6.93 \times 10^{-4}\ \mathrm{mol~L^{-1}} \times 162.11\ \mathrm{g~mol^{-1}} = 112.34 \times 10^{-4}\ \mathrm{g~L^{-1}} $$ The mass of water per liter: $$ 1\ \mathrm{L} \times 1\ \mathrm{kg~L^{-1}} = 1000\ \mathrm{g} $$ Now we calculate the mass fraction of calcium bicarbonate in the solution: $$ \mathrm{Mass\,Fraction} = \frac{112.34 \times 10^{-4}\ \mathrm{g}}{112.34 \times 10^{-4}\ \mathrm{g} + 1000\ \mathrm{g}} = 1.12 \times 10^{-3} $$ The maximum mass fraction of calcium bicarbonate in water at \(350\ \mathrm{K}\) (approximated) is about \(1.12 \times 10^{-3}.\)