Determination of a solubility product constant lab. Need help determining trail

Question

Determination of a solubility product constant lab. Need help determining trail 1 calculations Please show work so I understand O.o2o M Na, S,O A: Calcium Iodate, No Added Calcium Ion Concentration of Sodium Thiosulfate Solution (M) Volume of Calcium Iodate Solution (mL) Final Volume, Thiosulfate Solution (mL) Initial Volume, Thiosulfate Solution (mL) Volume of Thiosulfate Solution Added (mL) Moles Sodium Thiosulfate Used Moles Iodate Equilibrium Concentration of Iodate Ion (M) Equilibrium Concentration of Calcium Ion (M) Molar Solubility of Calcium Iodate(M) Trial #2 Trial #1 10.00 10.00 $632 4 2e 22 VKsp. Calcium lodate

Explanation / Answer

volume of the calcium iodate = 10 ml = 0.01 L

1. moles of iodate =0.001216 moles

equilibrium concentration = 0.001216 moles / 0.01 L =0.1216 M

2. as per standard reaction equilibrium concentration is half of the concentration of iodate

hence, concentration of calcium ion will be = 0.0608 M

3. solubility of calcium ion will be = 0.0608 M

4. solubility of iodate is 0.1216 M

and the value of Ksp = 0.0608 * (0.1216)2 = 8.99*10-6

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.