Question 19 of 28 IncorrectX ct General Chemistry 4th Edition niversity Science

Question

Question 19 of 28 IncorrectX ct General Chemistry 4th Edition niversity Science Books prenenited by Sapling Leaming Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.170 M pyridine, CsHsN(aq) with 0.170 M HBr(aq): Number (a) before addition of any HBr 9.23 Number (b) after addition of 12.5 mL of HBr 5.23 With equal concentrations of monoprotic titrant and analyte, the equivalence point would occur when the added volumes are equal (25.0 mL of HBr added). NumberWith (c) after addition of 17.0 mL of HBr 9.097 So at point (c), we are 17.0/50.0-68.0% of the way to the equivalence point. If 68.0% of the base has reacted, then 32.0% remains, and 68.0% of the conjugate base has been formed so [CSHENH ICsHsNH] 68.0/32.0. Number I (d) after addition of 25.0 mL of HBr 3.15 NumberFind the pOH using the Henderson-Hasselbalch equation and convert to pH for the final answer. (e) after addition of 37.0 mL of HBr4.75 C,H NH CsHsN pOH = pKb + log Previous Try Again (L) Next xit

Explanation / Answer

c)

mmoles of pyridine = 25 x 0.170 = 4.25

mmoles of HBr = 17 x 0.170 = 2.89

C5H5N   +   HBr   --------------> C5H5NH+

4.25           2.89                          0

1.36              0                            2.89

pOH = pKb + log [salt / base]

        = 8.77 + log [2.89 / 1.36]

pOH = 9.10

pH = 4.90

e)

mmoles of HBr = 37 x 0.170 = 6.29

here strong acid remains.

[H+] = 6.29 - 4.25 / 37 + 25 = 0.0329 M

pH = -log (0.0329)

pH = 1.48

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