Problem 1

A solution was prepared from 0.5284 g of an alloy containing aluminium. The aluminium was precipitated as aluminium 8-hydroxyquinolate,

\ce{Al(C9H6NO)3}

. The precipitate was separated, dissolved in hydrochloric acid, and the liberated 8-hydroxyquinoline was titrated with a standard potassium bromate solution containing potassium bromide; 17.40 cm

^3

of the standard solution were required, and the resultant product is a dibromo derivative of 8-hydroxyquinoline. The relative atomic mass of aluminium is 26.98. 1.1 Write the equation for the reaction of

\ce{Al^{3+}}

with 8-hydroxyquinoline. 1.2 Give the name of the type of compound formed in the precipitation. 1.3 Write the equation in which bromine is produced from bromate and bromide. 1.4 Write the equation for the reaction of bromine with 8-hydroxyquinoline. 1.5 Calculate the molar ratio of aluminium ions to bromate ions. 1.6 Calculate the percentage by weight of aluminium in the alloy.

Step 4 of 4: Percentage of aluminium

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\%\ce{Al} = \dfrac{17.40 \times 0.1000 \times 2.248}{528.4} \times 100 = 0.74\ \%

Analysis

The bromate titre in equivalents (17.40 cm $^3$ of 0.1000 N = $1.740\times 10^{-3}$ eq) times the equivalent mass of Al (2.248 mg meq $^{-1}$ ) gives $m(\ce{Al}) = 3.91$ mg; divided by the 528.4 mg sample this yields 0.74 % Al.