Calculate the pH for each of the following cases in the titration of 25.0 mL of

Question

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.230 M pyridine, CsH5N(aq) with 0.230 M HBr(aq): Number (a) before addition of any HBr 9.30 Number (b) after addition of 12.5 mL of HBr5.23 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) (c) after addition of 20.0 mL of HBr 4. So at point (c), we are 20.0/50.0 80.0% of the way to the equivalence point. If 80.0% of the base has reacted, then 20.0% remains, and 80.0% of the conjugate base has been formed so Number (d) after addition of 25.0 ml of HBr 3.09 HsNH'VCsH,NH) 80.0/20.0 NumberFind the pOH using the Henderson-Hasselbalch equation and convert to pH for the final answer (e) after addition of 36.0 mL of HBr 1.77 CHNH C,H,N O Previous ? Give Up & View Solution # Try Again O Next -el Exit

Explanation / Answer

c)

c) after the addition of 15 mL HBr

millimoles of acid = 20 x 0.230 = 4.6

millimoles of base = 25 x 0.230 = 5.75

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

5.75         4.6                                 0

1.15           0                                     4.6

pOH = pKb + log (4.6 / 1.15)

pOH = 9.37

pH = 4.63

e)

millimoles of HBr = 36 x 0.230 = 8.28

strong acid remained in the solution

[H+] = 2.53 / (25+36) = 0.0415 M

pH = -log(0.0415)

pH = 1.38

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