In a teaspoon (5.0 mL) of a liquid antacid, there are 400. mg of Mg(OH)2 and 400

Question

In a teaspoon (5.0 mL) of a liquid antacid, there are 400. mg of Mg(OH)2 and 400. mg of Al(OH)3. A 0.080 M HCl solution, which is similar to stomach acid, is used to neutralize 15.0 mL of the liquid antacid.

Part A What is the pH of the HCl solution?

Part B Write the balanced chemical equation for the neutralization of HCl and Mg(OH)2.Express your answer as a chemical equation. Identify all of the phases in your answer.

Part C Write the balanced chemical equation for the neutralization of HCl and Al(OH)3.Express your answer as a chemical equation. Identify all of the phases in your answer.

Part D How many milliliters of the HCl solution are needed to neutalize the Mg(OH)2?

Part E How many milliliters of the HCl sloution are needed to neutalize the Al(OH)3?

Explanation / Answer

(A)

HCl (aq.) -----------> H+ (aq.) + Cl- (aq.)

[H+] = [HCl] = 0.080 M

pH = - Log[H+] = - Log(0.080) = 1.10

(B)

2 HCl (aq.) + Mg(OH)2 (aq.) -----------> MgCl2 (aq.) + 2 H2O (l)

(c)

3 HCl (aq.) + Al(OH)3 (aq.) --------> AlCl3 (aq.) + 3 H2O (l)

(D)

Moles of Mg(OH)2 = mass / molar mass = 0.400 / 58 = 0.00690 mol

From the balanced equation,

Moles of HCl = 2 * moles of Mg(OH)2 = 2 * 0.00690 = 0.0138 mol

Volume of HCl needed = 0.0138 * 1000 / 0.080 = 172.4 mL

(E)

Moles of Al(OH)3 = mass / molar mass = 0.400 / 78 = 0.00513 mol

From the balanced equation,

Moles of HCl = 3 * moles of Al(OH)3 = 3 * 0.00513 = 0.0154 mol

Volume of HCl needed = 0.0154 * 1000 / 0.080 = 192.3 mL

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