Answer:
b. 0,08M NH₄⁺ / 0,12M NH₃
Explanation:
<em>The buffers are:</em>
<em>a. 0,08M H₂PO₄⁻ / 0,12M HPO₄²⁻</em>
<em>b. 0,08M NH₄⁺ / 0,12M NH₃</em>
It is possible to find out the pH of a buffer using Henderson-Hasselbalch formula:
<em>pH = pka + log₁₀ [A⁻] / [HA] </em><em>(1)</em>
Where A⁻ is the conjugate base of the weak acid, HA.
a. For the system H₂PO₄⁻ / HPO₄²⁻ pka is <u><em>7,198</em></u>. Replacing in (1)
pH = 7,198 + log₁₀ [0,12] / [0,08]
<em>pH = 7,37</em>
That means this buffer system don't have a pH of approximately 9,5
b. For the system NH₄⁺ / NH₃ pka is <em><u>9,3</u></em>. Replacing in (1)
pH = 9,3 + log₁₀ [0,12] / [0,08]
<em>pH = 9,50</em>
That means this buffer system is the buffer you need to use.
I hope it helps!
