Answer:
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]
