Question

Topic: Analytical Chemistry: Calculating solubilities at constant pH

Q: Calculate the molar solubility of Strontium Carbonate in a solution that has been fixed so that its pH is constant and equal to 6.00.

Answer 1

Answer on question #38477, Chemistry, Other

Calculate the molar solubility of Strontium Carbonate in a solution that has been fixed so that its pH is constant and equal to 6.00.

Solution:

Solubility of $\mathrm{SrCO_3}$ in pure water at $18^{\circ}C$ is equal 1,1 mg/100 ml or $7,45\cdot 10^{-5}$ M.

Acid dissociation constant of $\mathsf{H}_2\mathsf{CO}_3$: $\mathsf{pK}_{\mathrm{a1}} = 6,367$; $\mathsf{pK}_{\mathrm{a2}} = 10,32$.

Base dissociation constant of $\mathrm{Sr(OH)_2}$: $\mathsf{pK}_{\mathrm{b2}} = 0,82$.

$\mathrm{K_w = [H^+][OH^-]}$; $\mathrm{K_w = [H^+][OH^-]}$

In water next equilibriums are present:

$\mathrm{SrCO_3(s)} \rightleftharpoons \mathrm{Sr^{2+}(aq)} + \mathrm{CO_3^{2-}}$; solubility product: $\mathrm{K_{sp}} = [\mathrm{Sr^{2+}}][\mathrm{CO_3^{2-}}]$;

$\mathrm{CO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HCO}_{3}^{-} + \mathrm{OH}^{-}$; first hydrolysis constant $K_{1} = \frac{K_{w}}{K_{a2}} = \frac{[HCO_{3^{-}}][OH^{-}]}{[CO_{3^{2-}}]}$,

$\mathsf{pK}_1 = 3,68$

$\mathrm{HCO_3^- + H_2O \rightleftharpoons H_2CO_3 + OH^-}$; second hydrolysis constant

$$K _ {2} = \frac {K _ {w}}{K _ {a l}} = \frac {\left[ H _ {2} C O _ {3} \right] \left[ O H ^ {-} \right]}{\left[ H C O _ {3} \right]},$$

$\mathsf{pK}_2 = 7,633$

$\mathrm{Sr}^{2+} + \mathrm{H}_2\mathrm{O} \quad \mathrm{Sr(OH)}^+ + \mathrm{H}^+$; first hydrolysis constant $K_{1} = \frac{K_{w}}{K_{b2}} = \frac{[Sr(OH)^{+}][H^{+}]}{[Sr^{2+}]}$,

$\mathsf{pK}_1 = 11,18$

Second step of hydrolysis of $\mathrm{CO}_{3}^{2-}$ and first step of hydrolysis of $\mathrm{Sr}^{2+}$ are negligible at pH values near 7 (K₁/K₂ ≈ 10⁴) and can be ignored in further calculations.

In pure water $[\mathrm{Sr}^{2+}] = 7,45\cdot 10^{-5}\mathrm{M}$ and $[\mathrm{CO}_{3}^{2-}] + [\mathrm{HCO}_{3}^{-}] = 7,45\cdot 10^{-5}\mathrm{M}$.

Let $[\mathrm{HCO}_3^-] = [\mathrm{OH}^-]$ be equal to "y", then $[\mathrm{CO}_3^{2-}] = 7,45\cdot 10^{-5} - y$.

$K_{2} = \frac{y*y}{7,45\cdot 10^{-5} - y} = 2,09\cdot 10^{-4}$. Solving this equation we get value $y = 5,83\cdot 10^{-5}$.

So, $[\mathrm{HCO}_3^-] = 5,82\cdot 10^{-5}\mathrm{M}$ and $[\mathrm{CO}_3^{2-}] = 1,62\cdot 10^{-5}\mathrm{M}$.

Solubility product for $\mathrm{SrCO_3}$ is equal: $\mathrm{K_{sp}} = [\mathrm{Sr^{2+}}][\mathrm{CO_3^{2-}}] = 7,45\cdot 10^{-5}\mathrm{M} \cdot 1,62\cdot 10^{-5}\mathrm{M} = 1,2\cdot 10^{-9}\mathrm{M^2}$.

If the value of pH is constant molar fraction of different acid forms is depend on pH and don't depend on general concentration of acids forms. For dibasic acid $\mathrm{H}_2\mathrm{CO}_3$:

$$\phi \left(C O _ {3 ^ {2 -}}\right) = \frac {\left[ C O _ {3 ^ {2 -}} \right]}{\left[ C O _ {3 ^ {2 -}} \right] + \left[ H C O _ {3 ^ {-}} \right] + \left[ H _ {2} C O _ {3} \right]} = \frac {K _ {a l} K _ {a 2}}{K _ {a l} K _ {a 2} + K _ {a l} \left[ H ^ {+} \right] + \left[ H ^ {+} \right] ^ {2}};$$

If $\mathsf{pH} = 6$, $\phi (CO_{3^{2 -}}) = \frac{4,3\cdot 10^{-5}\cdot 4,79\cdot 10^{-11}}{4,3\cdot 10^{-5}\cdot 4,79\cdot 10^{-11} + 4,3\cdot 10^{-5}\cdot 10^{-6} + (10^{-6})^2} = 1,44\cdot 10^{-5}$;

So, $\mathrm{K_{sp}} = [\mathrm{Sr}^{2+}][\mathrm{CO}_{3}^{2-}]$; $[\mathrm{CO}_{3}^{2-}] = c(\mathrm{CO}_{3}^{2-}) \cdot \varphi(\mathrm{CO}_{3}^{2-}) = s \cdot \varphi(\mathrm{CO}_{3}^{2-})$; $[\mathrm{Sr}^{2+}] = s$.

$\mathrm{K_{sp}} = \mathrm{s} \cdot \mathrm{s} \cdot \varphi (\mathrm{CO}_{3}^{2-}) = 1,44 \cdot 10^{-5} \cdot \mathrm{s}^{2} = 1,2 \cdot 10^{-9}$. Hence, molar solubility of $\mathrm{SrCO_3}$ in solution with $\mathrm{pH} = 6$ is equal $\mathrm{s} = 9 \cdot 10^{-3} \mathrm{M}$.

Answer: $s = 9 \cdot 10^{-3} \mathrm{M}$