Question

Solid Al(NO3)3 is added to distilled water to produce a solution in which the concentration on nitrate, [NO3^-], is 0.10 M. What is the concentration of aluminum ion, [Al^3+] in this solution?

Answer 1

Answer:

The concentration of nitrate ion $NO_{3}^{-}$  is 0.033 M.

Explanation:

$Al(NO_{3})_{3} \rightarrow Al^{3+} + NO_{3}^{-}$

1 mole of $[Al^{3+}]$  = 3 mole $NO_{3}^{-}$

Let  $[NO_{3}^{-}]$  = 3y

then according to above reaction,

$[Al^{3+}]$ = y

$[NO_{3}^{-}]$ = 0.10 M (given)

$[Al^{3+}]$ = ?

3y = 0.10 M

$y = \frac{0.1}{3}$

y = 0.033 M

So ,concentration of $[Al^{3+}]$ = y = 0.033 M

Answer 2

Solid $Al(NO_3)_3$ is added to distilled water to produce a solution.

The Chemical equation:

$Al(NO_3)_3----->Al^{3+}+NO_3^-$

1 mole of aluminum ion produces 3 moles of nitrate ion.

Let the concentration of $[NO_3^-]$ be 3x.

So, from the given chemical equation:

$[Al^{3+}]=x$

Given:

$[NO_3^-]$ =0.01 M

To find:

$[Al^{3+}]=?$

$3x=0.01M\\\\x=\frac{0.01}{3} \\\\x=0.033M$

Thus, the concentration of aluminum ion, $[Al^{3+}]$ in this solution will be 0.033M.

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