Question

The solubility of barium iodate in an aqueous solution prepared by mixing 200 mL of 0.010 M barium nitrate with 100 mL of 0.10 M sodium iodate is $X \times 10^{-6} \, \text{mol dm}^{-3}$. \text{ The value of } $X$ \text{ is ------.}
Use: Solubility product constant $(K_{sp})$ of barium iodate = $1.58 \times 10^{-9}$

Hint:

In the presence of common ions, solubility decreases drastically. To find solubility in such a solution, use the initial ion concentrations and the solubility product expression carefully considering the common ion effect.

Answer 1

Step 1: Write the dissociation equation of barium iodate (\(\text{Ba(IO}_3)_2\)) in water: \[ \text{Ba(IO}_3)_2 \rightleftharpoons \text{Ba}^{2+} + 2 \text{IO}_3^- \] Step 2: Given, - Volume and concentration of barium nitrate \(\rightarrow V_1 = 200 \, \text{mL} = 0.200 \, \text{dm}^3\), \(C_1 = 0.010 \, \text{M}\) - Volume and concentration of sodium iodate \(\rightarrow V_2 = 100 \, \text{mL} = 0.100 \, \text{dm}^3\), \(C_2 = 0.10 \, \text{M}\) Step 3: Calculate the total volume after mixing: \[ V = V_1 + V_2 = 0.200 + 0.100 = 0.300 \, \text{dm}^3 \] Step 4: Calculate initial concentrations of ions from mixing before any precipitation: \[ [\text{Ba}^{2+}]_{\text{initial}} = \frac{C_1 \times V_1}{V} = \frac{0.010 \times 0.200}{0.300} = \frac{0.002}{0.300} = 0.00667 \, \text{M} \] \[ [\text{IO}_3^-]_{\text{initial}} = \frac{C_2 \times V_2}{V} = \frac{0.10 \times 0.100}{0.300} = \frac{0.010}{0.300} = 0.0333 \, \text{M} \] Step 5: Let the solubility of barium iodate in the mixed solution be \(S \, \text{M}\). At equilibrium: \[ [\text{Ba}^{2+}] = 0.00667 + S \] \[ [\text{IO}_3^-] = 0.0333 + 2S \] Since \(K_{sp}\) is very small, \(S\) will be very small compared to initial concentrations, so we approximate: \[ [\text{Ba}^{2+}] \approx 0.00667 \] \[ [\text{IO}_3^-] \approx 0.0333 \] Step 6: Calculate the ion product (Q) to check if precipitation occurs: \[ Q = [\text{Ba}^{2+}] [\text{IO}_3^-]^2 = (0.00667) \times (0.0333)^2 = 0.00667 \times 0.00111 = 7.4 \times 10^{-6} \] Since \(Q = 7.4 \times 10^{-6} \gg K_{sp} = 1.58 \times 10^{-9}\), the solution is supersaturated and precipitation will occur. Step 7: At equilibrium, \[ K_{sp} = [\text{Ba}^{2+}][\text{IO}_3^-]^2 = (0.00667 + S)(0.0333 + 2S)^2 \] Since \(S\) is small compared to initial concentrations, use approximation: \[ K_{sp} \approx (0.00667)(0.0333)^2 = 7.4 \times 10^{-6} \neq 1.58 \times 10^{-9} \] This contradiction implies we cannot neglect \(S\). Step 8: Define: \[ [\text{Ba}^{2+}] = 0.00667 + S \approx 0.00667 + S \] \[ [\text{IO}_3^-] = 0.0333 + 2S \] Write the equilibrium expression: \[ 1.58 \times 10^{-9} = (0.00667 + S)(0.0333 + 2S)^2 \] Since initial concentrations are much larger than \(S\), and \(K_{sp}\) is very small, precipitation will reduce ion concentrations to the saturation level. Step 9: For calculation of solubility \(S\), we consider only the amount of barium iodate that dissolves, ignoring initial concentrations. Solubility of barium iodate in pure water: \[ K_{sp} = s \times (2s)^2 = 4s^3 \] \[ s^3 = \frac{K_{sp}}{4} = \frac{1.58 \times 10^{-9}}{4} = 3.95 \times 10^{-10} \] \[ s = \sqrt[3]{3.95 \times 10^{-10}} \approx 7.36 \times 10^{-4} \, \text{M} \] Step 10: In presence of common ions, the solubility \(S\) decreases significantly. Approximate the effect by calculating the concentration from initial ions and use: \[ K_{sp} = (0.00667 + S)(0.0333 + 2S)^2 \approx (0.00667)(0.0333)^2 + \text{small correction terms} \] Given that \(K_{sp}\) is very small, the solubility \(S\) can be approximated by solving: \[ K_{sp} = (0.00667)(0.0333)^2 + \text{terms involving } S \] But since \(7.4 \times 10^{-6} \gg 1.58 \times 10^{-9}\), practically no more barium iodate will dissolve, so solubility \(S \approx 0\). Step 11: Therefore, solubility \(X \times 10^{-6} \approx 0\). Final Answer: \[ X \approx 0 \]