Answer:
The pH does not increase drastically because the NaOH reacts with the _HYDRAZOIC ACID__ present in the buffer solution.
Explanation:
NaOH reacts with the hydrazoic acid in the buffer
HN3 + OH? --> H2O + N3-
If you put acid in, it will react with N3-. Here is the equation below:
N3- + H+ --> HN3
A buffer contains 0.23 mol of hydrazoic acid and 0.27 mol of sodium azide in 1.00 L of solution. Upon the addition of 0.05 mol of NaOH, the pH does not increase drastically because the NaOH reacts with the hydrazoic acid present in the buffer solution.
A solution is prepared by dissolving 0.23 mol of hydrazoic acid (HN₃) and 0.27 mol of sodium azide (NaN₃) in water sufficient to yield 1.00 L of solution.
Hydrazoic acid is a weak acid since its Ka (1.9 . 10⁻⁵) << 1. The azide ion (N³⁻) coming from NaN₃ is its conjugate base. Together, they form a buffer system.
When a base is added, such as 0.05 moles of NaOH it reacts with the acid component and is neutralized by it.
NaOH + HN₃ ⇒ NaN₃ + H₂O
As a consequence of this neutralization, the pH does not increase drastically.
A buffer contains 0.23 mol of hydrazoic acid and 0.27 mol of sodium azide in 1.00 L of solution. Upon the addition of 0.05 mol of NaOH, the pH does not increase drastically because the NaOH reacts with the hydrazoic acid present in the buffer solution.
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