Map pu University Science Books presented by Sapling Learning eneral Chemistry 4
ID: 560724 • Letter: M
Question
Map pu University Science Books presented by Sapling Learning eneral Chemistry 4th Edition Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.160 M pyridine, CsHsN(aq) with 0.160 M HBr(aq): At point (a), this is a typical weak base question. Number 1321]| | Blaq)+H2O(l) BH+(aq) +0H-(aq) 13.21 (a) before addition of any HBr #0 + X 0.160 M Number 0.160 M-x (b) after addition of 12.5 mL of HBr 12.73 Number (c) after addition of 23.0 mL of HBr 11.82 Number 0.160 M-x 0.160 M where K 1.7 x10 and B(aq) is pyridine, (d) after addition of 25.0 mL of HBr 7 CsHsN(ag). Solve for x, which is equal to [OH then take the negative logarithm to find pOH which can be converted into pH for the answer to (a). Number (e) after addition of 36.0 mL of HBr 1.54 pH= 14.00-p(H
Explanation / Answer
millimoles of pyridine = 25 x 0.160 = 4
Kb= 1.7x10^-9
pKb = -logKb = -log (1.7x10^-9) = 8.77
a) before the addition of any HBr
pOH = 1/2 [pKb -logC]
= 1/2 [8.77 -log0.160]
= 4.78
pH + pOH = 14
pH = 9.22
b) after the addition of 12.5 mL HBr
it is half equivalence point
here salt millimoles = base millimoles
so pOH = pKb
pOH = 8.77
pH +pOH =14
pH = 5.23
c) after the addition of 16 mL HBr
millimoles of acid = 23 x 0.160 = 3.68
C6H5N + HBr ----------------------> C6H5NH+Br-
4 3.68 0
0.32 0 3.68
pOH = pKb + log (3.68 /0.32)
pOH = 9.83
pH = 4.17
d) after the addition of 25 mL HBr
it is equivalence point only salt is formed
salt concentration = millimoles / total volume = 4 / (25+25) = 0.08 M
salt is from strong acid weak base so pH <7
pH = 7 -1/2 [pKb + logC]
pH = 7 - 1/2 [8.77 + log 0.08]
pH = 3.16