An analytical chemist is titrating 1954 mL of a 02900 M solu

An analytical chemist is titrating 195.4 mL of a 0.2900 M solution of hydrazoic acid (HN;) with a 0.8300 M solution of NaOH. The pK, of hydrazolc aad is 4.72. Calculate the pH of the acid solution after the chemist has added 15.39 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.

Solution

It is titration of weak acid with a strong base.

HN3 + NaOH ------------------> NaN3 + H2O

195.4x0.29=56.666 15.39x 0.83=12.7737 0 0 initial mmoles

43.8923 0 12.7737 - after addition

The solution has a weak acid HN3 and its conjugate base NaN3.

Thus it bhaves as a buffer.

The pH is calculated using Hendersen equation.

pH = pKa + log [conjugae base]/[acid]

= 4.72 + log 12.7737/43.8923

= 4.1839

=4.18