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Consider the titration of 50.0 mL of 1.00 M C5H5N by 0.500 M HCl. For each volume of HCl added, decide which of the components is a major species after the HCl has reacted completely. Kb for C5H5N

User Bamba
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1 Answer

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The given question is incomplete. The complete question is as follows.

Consider the titration of 50.0 mL of 1.00 M
C_(5)H_(5)N by 0.500 M HCl. For each volume of HCl added, decide which of the components is a major species after the HCl has reacted completely. Kb for
C_(5)H_(5)N =
1.7 * 10^(-9). Calculate the pH at the equivalence point for this titration.

Step-by-step explanation:

Reaction equation of the given reaction is as follows.


C_(5)H_(5)N + H^(+) \rightarrow C_(5)H_(5)NH^(+)

Only
C_(5)H_(5)N is present in the solution before the addition of HCl. Hence, the equation will be as follows.


C_(5)H_(5)N + H_(2)O \overset{K_(b)}{\rightleftharpoons} C_(5)H_(5)NH^(+) + OH^(-)

And, at the equivalence point entire
C_(5)H_(5)N will completely react with the HCl. Hence, the solution contains
C_(5)H_(5)NH^(+) at the equivalence point. It is acidic with a pH less than 7.

No. of moles of HCl = No. of moles of
C_(5)H_(5)N


V_(HCl) * M_(HCl) = V_{C_(5)H_(5)N} * M_{C_(5)H_(5)N}


V_(HCl) * 0.5 M = 50 ml * 1 M


V_(HCl) = 100 ml

Hence, the volume of HCl at the equivalence point is 100 ml.

Also,
k_(a) = (k_(w))/(k_(b))

=
(1 * 10^(-14))/(1.7 * 10^(-19))

=
5.9 * 10^(-6)

Therefore, concentration of the acid is calculated as follows.

Concentration =
\frac{\text{No. of moles of acid}}{\text{Volume of solution}}

=
(0.05)/(0.15)

= 0.333 M

Since,
C_(5)H_(5)NH^(+) is a weak acid. So,


[H^(+)] = \sqrt{k_(a) * C}

=
\sqrt{5.9 * 10^(-6) * 0.333 M}

=
1.40 * 10^(-3)

Now, we will calculate the pH of the solution at the equivalence point as follows.

pH =
-log (1.40 * 10^(-3))

= 2.85

Thus, we can conclude that pH at the equivalence point for this titration is 2.85.

User RobinLovelace
by
7.9k points
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