Question

NONTON Aluminum ions may be precipitated from aqueous solution by the addition of hydroxide ion, forming Al(OH)3. A large excess of hydroxide ion must not be added, however, because the precipitate of Al(OH)3 will redissolve as a soluble compound containing aluminum ions, and hydroxide ions begin to form. How many grams of solid NaOH should be added to 10.0 mL of 0.131 M AICI: to just precipitate all the aluminum?

Answer 1

mass of NaOH = 0.157 g

Explanation

concentration AlCl₃ = 0.131 M

volume AlCl₃ = 10.0 mL = 0.0100 L

moles AlCl₃ = (concentration AlCl₃) * (volume AlCl₃ in Liter)

moles AlCl₃ = (0.131 M) * (0.0100 L)

moles AlCl₃ = 0.00131 mol

moles Al³⁺ = moles AlCl₃

moles Al³⁺ = 0.00131 mol

Al³⁺ (aq) + 3 OH^- (aq) $\rightarrow$ Al(OH)₃ (s)

moles OH^- = 3 * (moles Al³⁺)

moles OH^- = 3 * (0.00131 mol)

moles OH^- = 0.00393 mol

moles NaOH = moles OH^-

moles NaOH = 0.00393 mol

mass NaOH = (moles NaOH) * (molar mass NaOH)

mass NaOH = (0.00393 mol) * (40.0 g/mol)

mass NaOH = 0.1572 g