Question

just answer the first question but question is needed to solve

In a follow-up experiment, 19.494-mL of an aqueous solution of potassium hydroxide was prepared so that it contained exactly the same concentration of hydroxide ion from Question 10 This solution was mixed into a reaction vessel that contained an aqueous solution of a Group IIA-chloride chemical compound. At the end of the reaction, it was discovered that your solution was perfectly neutral.(Assume the hydroxide containing product precipitated out completely from the solution). How many ions of the Group IA cation must have been initially present in the reaction vessel? Input your response with 2 sig. figs.? Question 10 4 pts A neutralization experiment involving a commercially available antacid was performed according to the steps below: 1. Milk of Magnesia (antacid) is added to a beaker 2. An equivalent amount of a strong acid is added to the bearker 3. The reaction is allowed to proceed to completion Dosage amounts of active ingredients in a single dose (5-mL teaspoon) of antacid: Active Ingredients Dosage Amount aluminium hydroxide 400. mg magnesium hydroxide 400. mg Assuming the volume of your neutralization experiment in Step 1 was the recommended dosage of 4-teaspoon(s), what would be the concentration of hydroxide ion (in molarity) in the beaker? Input your response with 3 sig, figs. 4.37 M

Answer 1

Concentration of hydroxide ion = 5.82 M

Explanation

Volume of one teaspoon = 5 mL

Volume of 4-teaspoon = 4 * (5 mL) = 20 mL

total dose of aluminum hydroxide = 4 * (400 mg) = 1600 mg = 1.600 g

total dose of magnesium hydroxide = 4 * (400 mg) = 1600 mg = 1.600 g

moles aluminum hydroxide = (mass aluminum hydroxide) / (molar mass Al(OH)₃)

moles aluminum hydroxide = (1.600 g) / (78.0 g/mol)

moles aluminum hydroxide = 0.0205 mol

moles hydroxide = 3 * (moles aluminum hydroxide)

moles hydroxide = 3 * (0.0205 mol)

moles hydroxide = 0.0615 mol

moles magnesium hydroxide = (mass magnesium hydroxide) / (molar mass Mg(OH)₂)

moles magnesium hydroxide = (1.600 g) / (58.32 g/mol)

moles magnesium hydroxide = 0.0274 mol

moles hydroxide = 2 * (moles magnesium hydroxide)

moles hydroxide = 2 * (0.0274 mol)

moles hydroxide = 0.0549 mol

Total moles hydroxide = 0.0615 mol + 0.0549 mol

Total moles hydroxide = 0.116 mol

molarity hydroxide = (Total moles hydroxide) / (volume in Liters)

molarity hydroxide = (0.116 mol) / (20 x 10^-3 L)

molarity hydroxide = 5.82 M