Answer to Question #247263 in General Chemistry for banne

Question

Calculate the number of milliliters of 0.673 M Ba(OH)₂ required to precipitate all of the Al³⁺ ions in 111 mL of 0.662 M Al(NO₃)₃ solution as Al(OH)₃. The equation for the reaction is:

2Al(NO₃)₃(aq) + 3Ba(OH)₂(aq) 2Al(OH)₃(s) + 3Ba(NO₃)₂(aq)

Answer 1

We use the balanced equation (it was correctly balanced as it was presented) to find the volume of Ba(OH)₂ required by using conversion factors:

$111\text{ mL sol. }Al(NO_3)_3 \Big(\frac{0.662\text{ mol }Al(NO_3)_3)}{\text{1000 mL sol. }Al(NO_3)_3} \Big) \Big( \frac{3\text{ mol }Ba(OH)_2}{2\text{ mol }Al(NO_3)_3} \Big) \Big( \frac{\text{1000 mL sol. }Ba(OH)_2}{0.673\text{ mol }Ba(OH)_2} \Big) \\ \text{ } \\ = \text{163.778 mL} \approxeq \text{164 mL sol. }Ba(OH)_2 \text{ 0.662 M}$

In conclusion, we will need 164 mL of the Ba(OH)₂ solution to precipitate all the Al³⁺ ions as Al(OH)₃.

Reference

• Chang, R., & Goldsby, K. A. (2010). Chemistry. Chemistry, 10th ed.; McGraw-Hill Education: New York, NY, USA.