Question

Calcium oxalate $\left({\mathrm{CaC}}_2{\mathrm{O}}_4\right)$ is relatively insoluble in water $(K_{\mathrm{xp}}=$ $2\times {10}^{-9})$. However, calcium oxalate is more soluble in acidic solution. How much more soluble is calcium oxalate in $0.10\mathrm{M}$ ${\mathrm{H}}^{+}$ than in pure water? In pure water, ignore the basic properties of ${\mathrm{C}}_2{\mathrm{O}}_4^{2-}$

Short answer

Calcium oxalate is $1.42\times {10}^3$ times more soluble in a 0.10 M $H^{+}$ acidic solution than in pure water.

Step-by-step solution

Write solubility equilibria reactions for calcium oxalate in water and acid solution

First, we need to write the solubility equilibrium reactions for calcium oxalate in both water and acidic solution. For water: $Ca{C}_2{O}_4\rightleftharpoons C{a}^{2+}+C_2{O}_4^{2-}$ For acidic solution: In the presence of $H^{+}$ ions, one oxalate ion ($C_2{O}_4^{2-}$) will react with one $H^{+}$ ion to form hydrogen oxalate ($H{C}_2{O}_4^{-}$). $Ca{C}_2{O}_4+H^{+}\rightleftharpoons C{a}^{2+}+H{C}_2{O}_4^{-}$

Write solubility product expressions for both equilibria

Now, we will write the solubility product expressions for both equilibria. For water: $K_{sp}=[C{a}^{2+}][C_2{O}_4^{2-}]$ In this case, we can substitute x for the molar solubility of $Ca{C}_2{O}_4$: $$x\rightleftharpoons [C{a}^{2+}]+[C_2{O}_4^{2-}]$$ and $$K_{sp}=(x)(x)=x^2$$ For acidic solution: $$K_{sp}=[C{a}^{2+}][H{C}_2{O}_4^{-}]$$ In this situation, let y be the molar solubility of $Ca{C}_2{O}_4$ in the acidic solution. $$y\rightleftharpoons [C{a}^{2+}]=[H{C}_2{O}_4^{-}]$$ and $$K_{sp}=(y)(y)=y^2$$

Use the given $K_{sp}$ and concentration of $H^{+}$ ions to find solubility

The $K_{sp}$ of calcium oxalate in water is given as $2\times {10}^{-9}$. For water: $$x^2=2\times {10}^{-9}$$ Solve for x: $$x=\sqrt{2\times {10}^{-9}}$$ $$x=1.41\times {10}^{-5}$$ So, the molar solubility of calcium oxalate, x, in pure water is $1.41\times {10}^{-5}M$. For acidic solution: $$K_{sp}=(y)(0.10-y)$$ We can approximate that $y\ll 0.10$: $$K_{sp}\approx (y)(0.10)$$ Now, solve for y: $$y=\frac{K_{sp}}{0.10}$$ $$y=\frac{2\times {10}^{-9}}{0.10}$$ $$y=2\times {10}^{-8}$$ So, the molar solubility of calcium oxalate, y, in the acidic solution is $2\times {10}^{-8}M$.

Compare solubility in water and acidic solution

To find out how much more soluble calcium oxalate is in the acidic solution as compared to pure water, we need to find the ratio of solubility in the acidic solution to that in water. Solubility ratio: $$\frac{y}{x}=\frac{2\times {10}^{-8}}{1.41\times {10}^{-5}}$$ Calculating the ratio: $$\frac{y}{x}=1.42\times {10}^3$$ Thus, calcium oxalate is $1.42\times {10}^3$ times more soluble in the 0.10 M $H^{+}$ acidic solution than in pure water.

Key concepts

Calcium Oxalate Solubility

Calcium oxalate's solubility can often be puzzling due to its limited dissolution in water. Its chemical formula is ${\text{CaC}}_2{\text{O}}_4$, and it forms solid crystals, which are seldom soluble. However, dissolution occurs to a slight extent, breaking up into calcium ions $[{\text{Ca}}^{2+}]$ and oxalate ions $[{\text{C}}_2{\text{O}}_4^{2-}]$. This process is known as solubility equilibrium, where the solid and its dissolved ions exist in balance.

• In pure water, calcium oxalate's solubility product constant ($K_{sp}$) helps us understand the extent of its dissolution.
• The $K_{sp}$ quantifies the maximum concentration of ions that can co-exist in solution before saturation occurs and excess settles out as a solid.

However, introducing an acid, such as $0.10\text{ M}{\text{H}}^{+}$ changes the picture. The reaction shifts due to the formation of more soluble species, enhancing calcium oxalate's solubility.

Ksp (Solubility Product Constant)

The solubility product constant, $K_{sp}$, is an essential component in predicting how much of a sparingly soluble compound can dissolve in water. For calcium oxalate, with $K_{sp}=2\times {10}^{-9}$, this means that both the calcium and oxalate ion concentrations, when multiplied together, cannot exceed this value without forming a precipitate.The equation for its dissolution can be expressed as:$${\text{CaC}}_2{\text{O}}_4\rightleftharpoons {\text{Ca}}^{2+}+{\text{C}}_2{\text{O}}_4^{2-}$$Which translates into the solubility product expression:$$K_{sp}=[{\text{Ca}}^{2+}][{\text{C}}_2{\text{O}}_4^{2-}]$$This means in pure water, the concentration of both ions, represented as $x$, squared, must equal $2\times {10}^{-9}$.

• To find the concentration $x$, take the square root of the $K_{sp}$: $x=\sqrt{2\times {10}^{-9}}$.
• This handily reveals how soluble calcium oxalate is in water.

The $K_{sp}$ offers powerful insight and predicts solubility under varied environmental conditions.

Acidic Solution Solubility Enhancement

Increasing the solubility of calcium oxalate in an acidic environment involves fascinating chemical dynamics. When acids supply ${\text{H}}^{+}$ ions, reactions in the solution change the balance.In acidic solutions, oxalate ions become hydrogen oxalate by reacting with hydrogen ions:$${\text{C}}_2{\text{O}}_4^{2-}+{\text{H}}^{+}\to {\text{HC}}_2{\text{O}}_4^{-}$$This alters the original equilibrium, effectively removing ${\text{C}}_2{\text{O}}_4^{2-}$ from competition for calcium ions and increasing solubility. This process allows more calcium oxalate to dissolve, as seen in the enhanced molar solubility calculation:

• The new ion balance is solved using the same $K_{sp}$, substituting $[{\text{HC}}_2{\text{O}}_4^{-}]$ instead of $[{\text{C}}_2{\text{O}}_4^{2-}]$.
• The calculation approximates that $y$, the new molar solubility, must adjust to observed ${\text{H}}^{+}$ concentration influence.

This shifts solubility substantially, illustrated by an increase to $2\times {10}^{-8}\text{ M}$, showing acidic environments can notably boost solubility.