Answer to Question #154149 in General Chemistry for zelene

Question

An experiment was carried out as follows: 1.5766 g of an ammonium salt was placed in a beaker and 60 mL of 1.182 M sodium hydroxide was added. The resulting mixture was heated for about an hour and then left to cool to room temperature. The reaction mixture was then titrated with sulfuric acid and required 18.35 mL of the acid for complete neutralization. Given that the percentage by mass of ammonium in the salt was 37.5 %, calculate the molarity of the sulfuric acid used in the titration

Answer 1

Mass of aluminium in salt = 1.5766×35/100

0.5581 grams .

Aluminium salt + NaOH gives Al(OH)₃ .

Aluminium hydroxide will be titrated with Sulphuric acid.

MOLES of Al³⁺ = 0.5581/27= 0.0206 moles

No. Of equivalent = 0.206×3 = 0.0618 equivalent

No. Of equivalent used of sulphuric acid = 0.0618

No. Of moles = 0.0618/2 = 0.0309.

Molarity = 0.0305/0.0185 = 1.67 M