Respuesta :
The given question is incomplete. The complete question is as follows.
Consider the titration of 50.0 mL of 1.00 M [tex]C_{5}H_{5}N[/tex] 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 [tex]C_{5}H_{5}N[/tex] = [tex]1.7 \times 10^{-9}[/tex]. Calculate the pH at the equivalence point for this titration.
Explanation:
Reaction equation of the given reaction is as follows.
[tex]C_{5}H_{5}N + H^{+} \rightarrow C_{5}H_{5}NH^{+}[/tex]
Only [tex]C_{5}H_{5}N[/tex] is present in the solution before the addition of HCl. Hence, the equation will be as follows.
[tex]C_{5}H_{5}N + H_{2}O \overset{K_{b}}{\rightleftharpoons} C_{5}H_{5}NH^{+} + OH^{-}[/tex]
And, at the equivalence point entire [tex]C_{5}H_{5}N[/tex] will completely react with the HCl. Hence, the solution contains [tex]C_{5}H_{5}NH^{+}[/tex] at the equivalence point. It is acidic with a pH less than 7.
No. of moles of HCl = No. of moles of [tex]C_{5}H_{5}N[/tex]
[tex]V_{HCl} \times M_{HCl} = V_{C_{5}H_{5}N} \times M_{C_{5}H_{5}N}[/tex]
[tex]V_{HCl} \times 0.5 M = 50 ml \times 1 M[/tex]
[tex]V_{HCl}[/tex] = 100 ml
Hence, the volume of HCl at the equivalence point is 100 ml.
Also, [tex]k_{a} = \frac{k_{w}}{k_{b}}[/tex]
= [tex]\frac{1 \times 10^{-14}}{1.7 \times 10^{-19}}[/tex]
= [tex]5.9 \times 10^{-6}[/tex]
Therefore, concentration of the acid is calculated as follows.
Concentration = [tex]\frac{\text{No. of moles of acid}}{\text{Volume of solution}}[/tex]
= [tex]\frac{0.05}{0.15}[/tex]
= 0.333 M
Since, [tex]C_{5}H_{5}NH^{+}[/tex] is a weak acid. So,
[tex][H^{+}] = \sqrt{k_{a} \times C}[/tex]
= [tex]\sqrt{5.9 \times 10^{-6} \times 0.333 M}[/tex]
= [tex]1.40 \times 10^{-3}[/tex]
Now, we will calculate the pH of the solution at the equivalence point as follows.
pH = [tex]-log (1.40 \times 10^{-3})[/tex]
= 2.85
Thus, we can conclude that pH at the equivalence point for this titration is 2.85.
