A 25.0 mL sample of 0.100 M pyridine(K b = 1.70 x 10 - 9 ) is titrated withHCl.

Question

A 25.0 mL sample of 0.100 M pyridine(Kb = 1.70 x 10-9) is titrated withHCl.
Calculate the pH after the addition of 30.0 mL of 0.100 M HCl.

Explanation / Answer

Chemical reaction:                                   C5H5N(aq)   + HCl (aq)    ------------------>  C5H5NHCl (aq) Before rxn (moles)   0.025L*0.100M 0.030L * 0.1M                                             =0.0025            = 0.0030 After rxn(moles)                     0                  0.0005                                      0.0025                                [H+] = 0.0005 moles / 0.025 L + 0.03 L                                         = 0.0091 M                                pH    = - log [H+]                                        = 2.04 Here buffer is not forming as thenumber of moles of acid is more than base. So there will betwo acids after the reaction. One is strong acid another isconjugate acid of the weak base. As HCl is stronger than theconjugate acid, pH is affected by HCl. The pH will be same if we add HCl to pyridine and pyridine to HCl.There is no change if we reverse the addition of reactants. I hope it will surely help you. For further doubts let me know.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.