Question

A chemist titrates 90.0mL of a 0.3683M dimethylamine CH32NH solution with 0.8804M HNO3solution at

25°C. Calculate the pH at equivalence. The pKb of dimethylamine is 3.27. Round your answer to 2 decimal places.

A chemist titrates 90.0 mL of a 0.3683 M dimethylamine 1 solution with 0.8804 M HNO, solution at 25 °C. Calculate the pH at equivalence. The pK, of dimethylamine is 3.27. 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 HNO, solution added. pH = | x 5 ?

Answer 1

Molarity of (CH3)2NH = 0.368 M Volume of (CH3)2NH 90.00 ml = 0.090 L We know, Moles = Molarity (M)x Volume (V in L) Moles of (CH3)2NH = 0.3683 Mx 0.0900 L = 0.03315 mol At equivalent point initial moles of (CH3)2NH will be completely nutralized and same amount of (CH3)2NH2+ will be formed and Moles of Acid need to reach equivalent point = Moles of (CH3)2NH (since it is a mono protic acid) 0.0377 L Moles of HNO3 needed to reach equivalent = 0.03315 mol Molarity of HNO3 given = 0.8803 M Volume of HNO3 needed to reach equilibrium = Moles of Base / Molarity of Base 0.03315 mol = 0.8803 mol L-1 = 0.0377 L Total volume of solution at equivalent point = Volume of Acid + Volume of Base = 0.09000 L + = 0.1277 L ICE table for equivalent point can shown as below (CH3)2NH(aq) + H+(aq) <----->(CH3)2NH2+ (aq) Initial 0.03315 0.03315 0.0000 Change -0.03315 -0.03315 + 0.03315 Equilibrium 0.03315 Above ICE table shows only presence of (CH3)2NH2+, and moles of (CH3)2NH2+ = 0.03315 mol Concnetration of (CH3)2NH2+ = Moles of (CH3)2NH2+ / total volume = 0.03315 mol = 0.1277 L = 0.2597 M Now the pH of (CH3)2NH2+ can be calculated as shown below 0.0 H+ (aq) 0.0 Initial Change Equilibrium (CH3)2NH2+ (aq) <----->(CH3)2NH(aq) + 0.03315 0.0 (-x) (+x) (0.03315-x) (+x) . pKa= 14-pKb = 14- 3.27 = 10.73 Ka= 10^-10.73 = 1.86E-11 Acetate is a weak base and equilibrium constant Kb can be expressed as below Ka = ((CH3)2NHxH+)+((CH3)2NH2+) 1.86E-11 • (0.03315-x) so 1.86E-11 x^2 = 0.033 x^2 1.86E-11 0.0331 so x^2 6.165E-13 7.852E-07 but from ICE table x = [H+] so [H+] = 7.8520E-07 but pH = -log [H+] = -log [ 7.85197E-07 ] = asume x is small then (0.03315-x) will be 0.0331 pH of solution at equivalent point = 6.11