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CHM 131 Lecture Notes - Lecture 6: Barium Hydroxide, Sodium Hydroxide, Ideal Gas Law

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School
University of Rochester
Department
Chemistry
Course
CHM 131
Professor
Benjamin Hafensteiner

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Document Summary

The polarity of water will cause the polar solute to dissociate (break down). Nonpolar solutes will not be affected by the polarity of water. Mg has a +2 charge and so4 has a -2 charge. This means that there is one mg for every so4. So +2 and -2 will be in this solution. The acid is a diprotic acid (h2a), meaning that it has two h+ ions in the formula to donate to a base. The reaction is h2a(aq) + 2 naoh(aq) 2 h2o(l) + na2a(aq), where a2 is what is left over from the acid formula when the two protons (h+ ions) are reacted. For the hcl reaction, the base has the ability to accept two protons. The most common examples are ca(oh)2, sr(oh)2, and ba(oh)2. A possible reaction would be 2 hcl(aq) + ca(oh)2(aq) 2 h2o(l) + cacl2(aq).

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Related Questions

Show transcribed image text

Aluminum metal will reduce water, but the reaction is extremely slow under neutral conditions at room temperature.

Al(s) + 3H2O(l) → Al(OH)3(s) + H2(g) very, very slow

In the presence of aqueous base, however, Al will readily react to form the very soluble Al(OH)4–1(aq) complex anion which can be neutralized with H2SO4(aq) in two steps. The first produces the very insoluble Al(OH)3(s). If the Al(OH)3(s) slurry is then immediately boiled with additional H2SO4(aq) the insoluble hydroxide will react completely to give a clear, colorless solution. After excess water is boiled off, cooling produces KAl(SO4)2.12H2O(s), Alum, as fine, white needles.

Al(s) + 3H2O + KOH ⟶ KAl(OH)4(aq) + H2(g) fast and exothermic at room temperature

KAl(OH)4(aq) + ½ H2SO4(aq) ⟶ Al(OH)3(s) + H2O(l) + ½ K2SO4(aq)

Al(OH)3(s) + H2SO4(aq) ⟶ ½ Al2(SO4)3(aq) + 3 H2O (requires heating)

½ Al2(SO4)3(aq) + ½ K2SO4(aq) ⟶ KAl(SO4)2(s)

Combining these four equations gives the overall reaction.

Al(s) + KOH(aq) + 2 H2SO4(aq) ⟶ [KAl(SO4)2(aq)] + H2O(l) + H2(g) and

[KAl(SO4)2(aq)] + 12 H2O(l) ⟶ KAl(SO4)2.12H2O(s) (Alum)

A student dissolved 0.946g Al in 35mL of 2.00M KOH and obtained 13.8g of dry alum by following the experimental procedure given in this exercise. Assume that the Alum was 97.7% pure. Calculate the volume of 6.00 M H2SO4 (aq) needed just to precipitate Al(OH)3 (s).