Question

A chemist titrates 240.0 mL of a 0.7318 M ethylamine (C_2 H_5NH_2) solution with 0.3280 M HCl solution at 25 degree C. Calculate the pH at equivalence. The p^K_ b of ethylamine is 3.19. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HCl solution added.

Answer 1

5.73 is the answer

Explanation:

Here, moles of ethyl amine = 0.240 L x 0.7318 M = 0.175 moles

To reach equivalence point, 0.175 moles of HNO3 must be added

Therefore,

The volume of HNO3 that must be added = 0.177 moles / 0.3280 M = 535 mL

Total volume = 240 + 535 = 775 mL

Concentration of salt = 0.175 moles / 0.775 L = 0.225 M

We have,

pOH = $=(1 / 2)[\mathrm{p} \mathrm{E} \mathrm{w}+\mathrm{p} \mathrm{K} \mathrm{b}+\log \mathrm{C}]=(1 / 2)[14+3.19+\log 0.225]=8.27$

pH = 14 – pOH = 14-8.27 = 5.73

pH = 5.73