Consider the titration of 50.0 mL of 1.00 M C5H5N by 0.500 M HCI. For each volum

Question

Consider the titration of 50.0 mL of 1.00 M C5H5N by 0.500 M HCI. For each volume of HCI added, decide which of the components is a major species after the HCI has reacted completely Kb for CsHsN 1.7 x 10-9 added no: H yes: H2 no: CI yes: CsHsN no: C5HsNH+ HINT: Because no HCI has been added, this solution only contains a weak base and water. See text for discussion of weak base-strong acid titrations. yesH20 no CsHgN HINT: Assume the added strong acid neutralizes (reacts CsHsN CsHsNH+ CsHsN CSHsNH+ completely with) the weak base present. You have entered that Submit Answer Tries 0/45 Submit Answer Tries 0/45 You are correct. Your receipt no. is 168-8417 Previous Tries Submit Answer Incorrect. Tries 22/45 answer before Calculate the pH at the equivalence point for this titration. 2.85

Explanation / Answer

Moles of C5H5N taken = (volume of C5H5N in L)*(molarity of C5H5N) = (50.0 mL)*(1 L/1000 mL)*(0.1 M) = 0.05 mole.

30.00 mL of 0.500 M HCl is added:

HCl ionizes in water as below.

HCl (aq) ---------> H+ (aq) + Cl- (aq)

Moles of HCl = moles of Cl- = (30.0 mL)*(1 L/1000 mL)*(0.500 M) = 0.015 mole.

C5H5N reacts with H+ as below.

C5H5N (aq) + H+ (aq) --------> C5H5NH+ (aq)

Moles C5H5N reacted = moles H+ added = moles C5H5NH+ formed = 0.015 mole; moles C5H5N retained at equilibrium = (0.05 – 0.015) mole = 0.035 mole.

H+: no

H2O: yes

Cl-: yes

C5H5N: yes

C5H5NH+: no

100.00 mL of 0.500 M HCl is added:

Moles of HCl = moles of Cl- = (100.0 mL)*(1 L/1000 mL)*(0.500 M) = 0.05 mole.

Moles C5H5N reacted = moles H+ added = moles C5H5NH+ formed = 0.05 mole; moles C5H5N retained at equilibrium = (0.05 – 0.05) mole = 0.00 mole.

H+: no

H2O: yes

Cl-: yes

C5H5N: no

C5H5NH+: yes

150.00 mL of 0.500 M HCl is added:

Moles of HCl = moles of Cl- = (150.0 mL)*(1 L/1000 mL)*(0.500 M) = 0.075 mole.

C5H5N is completely neutralized in the last step and the solution now contains C5H5HN+ (a weak base) and excess HCl (a strong acid); therefore, we have

H+: yes

H2O: no

Cl-: yes

C5H5N: no

C5H5NH+: yes

The equivalence point is after the addition of 100.00 mL of 0.500 M HCl (the 2nd step above) when C5H5N is completely neutralized to give C5H5NH+. C5H5NH+ is a weak acid and re-establishes equilibrium as

C5H5NH+ (aq) ---------> C5H5N (aq) + H+ (aq)

[C5H5NH+] = (moles C5H5NH+ formed)/(volume of solution in L) = (0.05 mole)/[(130.00 mL)*(1 L/1000 mL)] = 0.3846 M.

Since H+ is formed, we must work with Ka, the acid ionization constant. Given Kb, we must have

Ka = Kw/Kb where Kw = 1.0*10-14 is the ionic product of water.

====> Ka = (1.0*10-14)/(1.7*10-9) = 5.88*10-6

Ka = [C5H5N][H+]/[C5H5NH+] = (x).(x)/(0.3846 – x)

Since Ka is small, we have, (0.3846 – x) M 0.3846 M; therefore,

5.88*10-6 = x2/(0.3846)

=====> x2 = 2.26*10-6

Therefore, [H+] = 2.26*10-6 M and pH = -log [H+] = -log (2.26*10-6) = 5.64589 5.64 (ans).

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