Calculate the pH for each of the following cases in the titration of 25.0 mL of
ID: 1033947 • Letter: C
Question
Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.100 M pyridine, CsHsN(aq) with 0.100 M HBr(aq): Number (a) before addition of any HBr 9.115 Number (b) after addition of 12.5 mL of HBr 5.23 th equal concentrations of monoprotic titrant and analyte, the equivalence point would occur when the added volumes are equal (25.0 mL of HBr added) Number (c) after addition of 24.0 mL of HBr 4.65 So at point(c), we are 24.0/50.0 96.0% of the way to the equivalence point. If ga0% of the base has reacted, then 4.0% remains, and 96.0% of the conjugate base has been formed so CsHsNH VICsH&NH;)-96.0/4.0 (d) after addition of 25.0 mL of HBr 3.28 Number Find the pOH using the Henderson-Hasselbalch equation and convert to pH for the final answer. (e) after addition of 36.0 mL of HBr 1.744 CH NH CSH,N pOH-pK,+ logExplanation / Answer
Before any addition of HBr,
pOH = 1/2(pKb - log C)
pOH = 1/2(8.77 - log(25*0.1/1000))
pOH = 5.686
pH = 14 - 5.686 = 8.314
After addition of 12.5 mL of HBr,
pOH = pKb + log([C5H5NH+]/[C5H5N])
pOH = 8.77 + log((0.1*12.5)/((0.1*25)-(0.1*12.5)))
pOH = 8.77
pH = 14 - 8.77 = 5.23
After addition of 24 mL of HBr,
pOH = 8.77 + log((0.1*24)/(0.1*25-0.1*24))
pOH = 8.77 + 1.38 = 10.45
pH = 14 - 10.45 = 3.55
After addition of 25 mL of HBr,
number of moles of HBr = number of moles of C5H5N
pH = 7
After addition of 36 mL of HBr,
number of moles of HBr = 0.1 *36 = 3.6 mmol
number of moles of Pyridine = 0.1 * 25 = 2.5 mmol
number of moles of HBr remaining = 3.6 - 2.5 = 1.1 mmol
pH = -log(1.1*10^-3/(36+25)*1000)
pH = -log(0.01803)
pH = 1.744