Question

Write the equations for the reactions of each compound in water which leads to the solution to be acidic or basic:

0.1M NaH2PO4

0.1M HC2H3O2

0.1M ZnSO4

0.1M Na2CO3

0.1M NaC2H3O2

0.1M NaHSO4

0.1M NaOH

Answer 1

1. NaH₂PO₄(s) → Na⁺(aq) + H₂PO₄⁻(aq)

   The H₂PO₄⁻ ion is amphoteric. It can act as either an acid or a base.The equation for its reaction as an acid is

     H₂PO₄⁻ + H₂O ⇌ HPO₄²⁻ + H₃O⁺; pKa=7.21

    The equation for its reaction as a base is

     H₂PO₄⁻ + H₂O ⇌ H₃PO₄ + OH⁻; pKb=11.88

    Since pKa<pKb, H₂PO₄⁻ is a stronger acid than it is a base.

    This means that solutions of NaH₂PO₄ will be acidic

2. CH₃COOH(aq)+H₂O(l)⇌H₃O⁺(aq)+CH₃COO⁻(aq)

3.ZnSO_⁴ is a salt of Strong acid H₂SO₄ and weak base Zn(OH)₂. So it is acidic in nature

    ZnSO₄ + H₂OZn²⁺ + SO₄^2- +H₂O

Out of which Zn²⁺ is active ion . so it react with water as

Zn²⁺+2H₂OZn(OH)₂+2H⁺

4. Na₂CO₃ (s) —-> 2 Na⁺(aq) + CO₃^–2(aq)

   

   The carbonate ion is able to remove protons (H⁺) from water to form bicarbonate ions and hydroxide ions.      

   Hence it is the conjugate base of the bicarbonate ion;

   CO₃^–2(aq) + H₂O(l) ←-> HCO₃^-(aq) + OH^-(aq)

   The overall increase in OH^- ions due to these reactions results in an increase in the pH of the solution, and cause it to become alkaline.

5. NaC₂H₃O₂ is a basic salt because the acetate ion will pull an H⁺ ion from water to form the weak acid CH₃COOH (Acetic acid). Since acetic acid is weak, it will mostly remain in water as is which will leave some OH^- ions floating around. The OH^- ions floating around makes the solution basic.

       CH₃COONa + H₂OCH₃COOH+Na⁺+OH^-

^6. NaHSO₄ is a very weak acid, in water it will dissociate into Na⁺ and HSO₄^- ions and HSO₄^- can give off some protons since it acts as a weak acid.

NaHSO₄ +H₂ONa⁺+HSO₄^-+H₂O

HSO^4-+H₂OSO₄^2-+H₃O⁺

7. NaOH+H₂ONa⁺+H₃O⁺