A student places 25.0 mL of 0.100 M hydrazoic acid (HN3) in an Erlenmeyer flask and adds indicator. She then adds 5.0 mL of standardized 0.200 M NaOH. What is the pH of the solution

pH = 4.42

Explanation:

HN3, a weak acid reacts with NaOH as follows:

HN3 + NaOH  → H2O + NaN3

Where 1mol HN3 reacts with 1mol NaOH

After the reaction of only a part of HN3, you will have in solution HN3 (Weak acid) and NaN3 (Conjugate base). This mixture produce a buffer that follows H-H equation:

pH = pKa + log [Conjugate base] / [Weak acid)

Where pH is the pH of the buffer

The hydrazoic acid is a weak acid with pKa = 4.6

And [] could be taken as the moles of each species

After the reaction, the moles of NaOH added = Moles NaN3 produced

And moles HN3 = Initial moles HN3 - Moles NaOH

Moles NaOH = Moles NaN3:

5x10⁻³L * (0.200mol / L) = 1x10⁻³ mol NaN3

Initial moles HN3:

0.025L * (0.100mol / L) = 2.5x10⁻³ moles HN3

final moles: 2.5x10⁻³ moles HN3 - 1x10⁻³ mol = 1.5x10⁻³ moles HN3

Replacing in H-H equation is:

pH = 4.6 + log [1x10⁻³ mol NaN3] / [1.5x10⁻³ moles HN3]