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3.1. complete neutralization requires one H⁺ for each OH^-.

Al(OH)₃ has three avaliable OH^- ions and H₂SO₄ can only provide two H⁺ ions, the re action requires two moles of Al(OH)₃ for three moles of H₂SO₄.

2Al(OH)₃ + 3 H₂SO₄ -------> Al₂(SO₄)₃ + 6H₂O

3.2. Molecular: H₃PO₄(aq) + Ca(OH)_2(aq) ----------> CaHPO₄(aq) + 2H₂O(l)

Total ionic : H₃PO₄(aq) + Ca²⁺(aq) + 2OH^-(aq) ---------> Ca²⁺(aq) + HPO₄^2-(aq) + 2H₂O (l)

Net Ionic : H₃PO₄(aq) + 2OH^- -------> HPO₄^2-(aq) +2H₂O(l)

3.3.a. HNO₃ + NaOH -------> NaNO₃ + H₂O

3.3.b. Ca(OH)₂ + 2 HI -------> CaI₂ + 2H₂O

3.3.c. HCLO₂ + KOH -------> KClO₂ + H₂O

3.4. Percent Ionization of HF = (concentration ionized/ original concentraion) X 100%

= ([0.05]/1.0 M) X 100% = 5%

3.5. H₂S + LiOH -----> LiHS + H₂O

LiHS + LiOH -----> Li₂S + H₂O

3.6. H₂S(aq) + 2 NaOH (aq) ---------> Li₂S(aq) + 2H₂O(l)