Question

(2 points) A 50.0 mL solution of 0.040 M pyridine (C6HSN) was titrated with 0.0500 M HCl. Calculate the pH at the following volumes of added acid: Va-0, 15, Ve, 30.0, Ve, and 45.0 mL. Kb 1.7x109 2.

Answer 1

Answer:

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an pH at the initial point (vo): Constructing CGHSN - CEHSNH toh I.C.E table K6 = 17X10-9 pka = -dog (ko) = 877 Initid Dolom Om om Change - - tx tx Equi. 0.040-X Ko= aix hool - o n x2 +1.7 X10-9% -6.8X10"-0 This will gives X 8.2453 X10M = [OH-] pOH = -log (8-2453x10 pH= 14- pOH = 8.92 (at vo) = 5.08 At V₂ = 15 mL moles of H+ = 10.05/1000) x 15 = 7.5×16-4 moles Total volume = sotis = 65 mL or 0.065 [C6Hs N]. = 0.04 m contains 2x10-3 moles in 50 mL solution. Now we will use the Henderson Equation: pOH = pk + log LBHt/Bot] Addition of Hol will neutenlize the pyridine and produce conjugate base. HN] = ( 2x lo-3-75 X 60°°) - D1 D142M, C, H,NH+] = 75 x 10 = 0.0 4 M 0.065 0.065 pOH = 8.77 + log (0:01154 ) = 8.55 pH= 14-8.55 = 5.45 0.0192 At ve pOH = pkl = 8.77 col- ok. = 8.77 and pH = 14-8.77 = ano 5.23 At va = 3 om moles of H* = 0.05/1000 x 30 = 15 x 10-3 total volume = 80 mL = 0.08L [CGHSN] = 6.25X10-3M [CGH SNH+] = 0.01875 M. pOH = 8.77 + log/0.01875 = 9.25 pH = 4.75 625*10-3) Atve we only have C6H5NH * species and ka= kw/kb = 10-14/1.7x10-9 = 5.888106 total volume = 50+ 40 = 90 mL = 0.09 L and concentration [C6H, NH+] = 0.02222 m. and expression for ka= x² or x² + 5.88% 10 x - 1.289 x 10 =0 D'02222-x) 2 = 0.00035609 = +) and pH = -log (0.00035609) = 3:49 At 45 mL Va Total volume = 95 mL and pt governs by the Her After ve Extra amount of SML He will produce = 2.5X 16-4 moles H+]æ p.5X104 - 2.632 x 10-3 m and pha-log (2.632 X10-3) = 2:58 0.095

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