Question

I've been trying to do this, and I keep getting the wrong answer!

A 28.3 mL sample of 0.239 M trimethylamine, (CH₃)₃N, is titrated with 0.328 M hydroiodic acid.

After adding 8.10 mL of hydroiodic acid, the pH is.....

Answer 1

Molarity of (CH3)3N Volume of (CH3)3N Moles of acid or Base Moles of (CH3)3N 0.239 M 28.30 ml * (0.001 L/1.0 mL) 0.0283 L Molarity (M) Volume (V in L) 0.239 mol/L * 0.0283 L 0.0068 mol 0.328 M Molarity of HI Volume of HI added 8.100 ml * (0.001 L/1.0 mL) 0.0081 L 0.3280 mol/L * 0.0081 L 0.0027 mol Moles of HI Kb of (CH3)3N 5.20E-04 pKb = -log Kb = -log [ 5.20E-04 ] = 3.28 pka 14 - pka = 10.72 When Acid is added to the buffer, it reacts with the Base component ((CH3)3N) and produce its conjugate Acid (CH3)3NH+). So during the addition of strong acid to the buffer, (CH3)3N moles are decreased and (CH3)3NH+ moles are increased as shown in the below table. The Net ionic equation after addition of HI is (CH3)3N (aq) + Initial 0.0068 Change -0.00266 Equilibrium 0.0041 Henderson equation for buffer solution is H+ (aq) 0.00266 -0.00266 0.000 <-------> (CH3)3NH+ (aq) 0.0000 + 0.00266 0.00266 рн pka + r Conj. Base log L Acid Here the Base is (CH3)3N and Acid is (CH3)3NH+ pH + 10.72 = 0.00411 log L 0.00266 log [ 1.5458 ] 0.1892 = 10.905 + 10.72 10.72 + pH of solution 10.9