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Consider the titration of a 28.0 −mL sample of 0.170 MCH3NH2 with 0.150 M HBr. Kb for methylamine, CH3NH2, at 25 C is 4.4 x 10-4 Determine each of the following. |
the pH at 6.0 mL of added acid
the pH at one-half of the equivalence point
the pH at the equivalence point
the pH after adding 6.0 mL of acid beyond the equivalence point
1)
mmoles of CH3NH2 = 28 x 0.170 = 4.76
mmoles of HBr = 6 x 0.150 = 0.90
pKb = 3.356
pKa = 10.644
CH3NH2 + HBr ------------> CH3NH3+
4.76 0.90 0
3.86 0 0.90
pH = pKa + log [base / acid]
= 10.644 + log [3.86 / 0.90]
pH = 11.28
2)
At half - equivalence point :
pH = pKa
pH = 10.64
3)
At equivalence point :
volume of acid = 31.73 mL
here only salt remains .
salt concentration = 4.76 / 31.73 + 28 = 0.0797 M
pH = 7 - 1/2 (pKb + log C)
= 7 - 1/2 (3.356 + log 0.0797)
pH = 5.87
4)
mmoles of acid = 37.73 x 0.150 = 5.66
[H+] = 5.66 - 4.76 / 37.73 + 28 = 0.0137 M
pH = -log (0.0137)
pH = 1.86
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