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