Question

An analytical chemist is titrating 199.0 mL of a 1.0000 Msolution of hydrazoic acid (HN) with a 0.3400 Msolution of NaOH. The p K of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 164.8 mL of the NaOH solution to it Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added. Round your answer to 2 decimal places. pH ?

Answer 1

use:

pKa = -log Ka

4.72 = -log Ka

Ka = 1.905*10^-5

Given:

M(HN3) = 1 M

V(HN3) = 199 mL

M(NaOH) = 0.34 M

V(NaOH) = 164.8 mL

mol(HN3) = M(HN3) * V(HN3)

mol(HN3) = 1 M * 199 mL = 199 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.34 M * 164.8 mL = 56.032 mmol

We have:

mol(HN3) = 199 mmol

mol(NaOH) = 56.032 mmol

56.032 mmol of both will react

excess HN3 remaining = 142.968 mmol

Volume of Solution = 199 + 164.8 = 363.8 mL

[HN3] = 142.968 mmol/363.8 mL = 0.393M

[N3-] = 56.032/363.8 = 0.154M

They form acidic buffer

acid is HN3

conjugate base is N3-

Ka = 1.905*10^-5

pKa = - log (Ka)

= - log(1.905*10^-5)

= 4.72

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.72+ log {0.154/0.393}

= 4.313

Answer: 4.31