Question

An analytical chemist is titrating 184.7 mL of a 0.5900 M solution of ethylamine (C2H3NH2) with a 0.1800 M solution of HIO3. The pK, of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 712.2 mL of the HIO, solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HIO, solution added. Round your answer to 2 decimal places. pH = 0 1 x 5 ?

Answer 1

Ans :-

Molarity of ethylamine = 0.5900 M and Volume of solution = 184.7 mL = 0.1847 L

Therefore,

No.of moles of ethylamine = Molarity of ethylamine x Volume of solution in L = 0.5900 M x 0.1847 L

= 0.108973 mol

Also, Molarity of HIO₃ = 0.1800 M and Volume of solution = 712.2 mL = 0.7122 L

No.of moles of HIO3 = Molarity x Volume in L = 0.1800 M x 0.7122 L

= 0.128196 mol

ICF table is :

.................................(C₂H₅)NH₂ (aq).............+..............H⁺ (aq) <---------------> (C₂H₅)NH₃⁺ (aq)

Initial .........................0.108973 mol.............................0.128196 mol ...................0.0 mol

Change.....................-0.108973 mol..............................-0.108973 mol...................+0.108973 mol

Final...........................0.0 mol.........................................0.019223 mol........................0.108973 mol

pH only depends upon HIO₃ as it is more acidic than (C₂H₅)NH₃⁺

Therefore,

[H⁺] = No. of moles of HIO₃/Volume of solution in L = 0.019223 mol / 0.8969 L = 0.02143 mol/L

pH = - log [H⁺] = - log 0.02143 M = 1.67

Therefore, pH of the solution = 1.67