In this problem we need to identify the relative molar amounts of:
NH₄⁺ , ClO₄⁻ , NH₃ , HClO₄ , H₃O⁺ , H₂O, OH⁻
- The salt is completely ionized through this equation:
NH₄ClO₄ → NH₄⁺ + ClO₄⁻
0.01 M 0.01 M 0.01 M
So:
[ClO₄⁻] = 0.01 M (conjugate base of strong acid)
[HClO₄] = 0 M
Note that: NH₄⁺ is weak acid ionized as follow:
NH₄⁺ ⇄ NH₃ + H⁺
Ka = [tex] \frac{[ H^{+}][ NH_{3}]}{ [NH_{4} ^{+}] } [/tex] = 5.6 x 10⁻¹⁰
[H⁺] = [NH₃] = x and [NH₄⁺] = 0.01 - x (because Ka is very small so x will be neglected compared to 0.01)
Ka = 5.6 x 10⁻¹⁰ = [tex] \frac{ x^{2} }{0.01} [/tex]
[H⁺] = [NH₃] = x = 2.37 x 10⁻⁶ M
[NH₄⁺] = 0.01 M
[OH⁻] = [tex] \frac{1 X 10^{-14} }{[H^{+} ]} [/tex] = 4.22 X 10⁻⁹ M
because water is the solvent & this is very dilute solution:
[H₂O] = [tex] \frac{1000g/L}{18 g/mol} [/tex] = 55.6 M H₂O
