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Math  /  Algebra

QuestionEXAMPLE 7-5 Calculate the solubility of Ba(IO3)2\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2} in a solution prepared by mixing 200 mL of 0.0100MBa(NO3)20.0100 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} with 100 mL of 0.100 M NaIO .

Studdy Solution

STEP 1

1. The solubility product constant (Ksp K_{sp} ) for Ba(IO3)2\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2} is known or can be found.
2. The solution is mixed, and the volumes are additive.
3. The reaction reaches equilibrium.
4. The common ion effect will influence the solubility of Ba(IO3)2\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}.

STEP 2

1. Determine the initial concentrations of ions in the mixed solution.
2. Set up the equilibrium expression using the solubility product constant.
3. Solve for the solubility of Ba(IO3)2\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}.

STEP 3

Calculate the total volume of the mixed solution: Vtotal=200 mL+100 mL=300 mL V_{\text{total}} = 200 \, \text{mL} + 100 \, \text{mL} = 300 \, \text{mL}
Calculate the initial concentration of Ba2+\mathrm{Ba}^{2+} from Ba(NO3)2\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}: [Ba2+]initial=0.0100 M×200 mL300 mL=0.00667 M [\mathrm{Ba}^{2+}]_{\text{initial}} = \frac{0.0100 \, \text{M} \times 200 \, \text{mL}}{300 \, \text{mL}} = 0.00667 \, \text{M}
Calculate the initial concentration of IO3−\mathrm{IO}_{3}^{-} from NaIO3\mathrm{NaIO}_{3}: [IO3−]initial=0.100 M×100 mL300 mL=0.0333 M [\mathrm{IO}_{3}^{-}]_{\text{initial}} = \frac{0.100 \, \text{M} \times 100 \, \text{mL}}{300 \, \text{mL}} = 0.0333 \, \text{M}

STEP 4

Write the dissolution equation for Ba(IO3)2\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}: Ba(IO3)2(s)⇌Ba2+(aq)+2IO3−(aq) \mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2} (s) \rightleftharpoons \mathrm{Ba}^{2+} (aq) + 2 \mathrm{IO}_{3}^{-} (aq)
Write the expression for the solubility product constant (Ksp K_{sp} ): Ksp=[Ba2+][IO3−]2 K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{IO}_{3}^{-}]^2

STEP 5

Let s s be the solubility of Ba(IO3)2\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2} in mol/L. At equilibrium: [Ba2+]=0.00667+s [\mathrm{Ba}^{2+}] = 0.00667 + s [IO3−]=0.0333+2s [\mathrm{IO}_{3}^{-}] = 0.0333 + 2s
Assuming s s is small, approximate: [Ba2+]≈0.00667 [\mathrm{Ba}^{2+}] \approx 0.00667 [IO3−]≈0.0333 [\mathrm{IO}_{3}^{-}] \approx 0.0333
Substitute into the Ksp K_{sp} expression: Ksp=(0.00667)(0.0333)2 K_{sp} = (0.00667)(0.0333)^2
Solve for s s using the known Ksp K_{sp} .
The calculated solubility s s will be the solubility of Ba(IO3)2\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2} in the mixed solution.

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