Consider the titration of 100.0 mL of 0.100 M HONH2 by 0.200 M HCl. Kb for HONH2

Question

Consider the titration of 100.0 mL of 0.100 M HONH2 by 0.200 M HCl. Kb for HONH2 = 1.1 x 10-8. The pH at the equivalence point for this titration is 3.61. At what volume of HCl added in this titration does the pH = 6.04? Express your answer in mL and include the units in your answer. HINT: Determine the pKb for the weak base, then determine the pKa for the conjugate acid and compare the pKa value to the pH. See text for discussion of weak base-strong acid titrations.

Explanation / Answer

HONH2 + HCl ---> HONH3+ + Cl- at the equivalence point n acid = n base = M*V = 0.100*0.100 = 1.00*10^-2 mol V acid added = n / M HCl = 1.00*10^-2 / 0.200 = 5.00*10^-2 L = 50.0 mL total volume at EP = 0.100 L + 5.00*10^-2 L = 0.150 L [HONH3+]= n / V = 1.00*10^-2 / 0.150 = 6.67*10^-2 M the solution will be acidic [H+] = square root (6.67*10^-2*Kw/Kb) = 2.46*10^-4 M pH = 3.61 pKa of HONH3+ is 6.04 pH=pKa means half of the equivalence point that means V HCl = 25.0 mL

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