1. Anti-acid tablets contain CaCO3 as the active ingredient. Write the balanced

Question

1. Anti-acid tablets contain CaCO3 as the active ingredient. Write the balanced equation for the reactivity of CaCO3 with HCl. 2. Calculate the molar solubility of CaCO3. Ksp for CaCO3 is 5.0 x 10-3. 3. Calculate the pH of CaCO3 solution. 4. If you dissolve CaCO3 in pH 5 solution what would you expect for the molar solubility? It will be greater or smaller. 5. If you dissolve CaCO3 in pH 8 solution what would you expect for the molar solubility? It will be greater or smaller.

Explanation / Answer

A 0.250 g antacid tablet containing CaCO3 as the active ingredient required 25.82 mL of 0.115 M HCl for? complete neutralization. What was the mass percent of CaCO3 in the tablet? First, start with the balanced reaction between CaCO3 and HCl: CaCO3(s) + 2HCl(aq) ---> CaCl2(aq) + H2O(l) + CO2(g) Next, convert milliliters of HCl to liters of HCl, by dividing the number of milliliters by a thousand 25.82mL / 1000 = .02582 L Multiply the liters of acid by the molarity of acid in order to calculate the moles of acid needed for neutralization. .02582L * .115M = .00297 moles of HCl From the balanced equation, you can see that the moles of CaCO3(s) is half the moles of HCl. Multiply .00297 by a half to figure out the moles of CaCO3(s). .00297 / 2 = .00148 moles of CaCO3(s) Multiply the moles of CaCO3(s) by the molar mass of CaCO3(s), which is 100.0869 g/mole. .00148 moles * 100.0869 g/mole = .148 grams of CaCO3(s) Mass percent is grams of CaCO3(s) divided by the total mass and then multiplied by 100%. 100% * [grams of CaCO3(s) ] / [total mass] 100% * [.148 grams ] / [0.250 grams] = 59.2%

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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