Complete Question
Calculate the pH during the titration of 20.00 mL of 0.1000 M trimethylamine, (CH3)3N(aq), with 0.2000 M HClO4(aq) after 9.48 mL of the acid have been added.Kb of trimethylamine = 6.5 x 10-5.
Answer:
The pH is [tex]pH = 9.84[/tex]
Explanation:
From the question we are told that
The volume of trimethylamine, (CH3)3N(aq) is [tex]V_{t} = 20.00mL[/tex]
The concentration of trimethylamine is [tex]C_t = 0.1000 \ M[/tex]
The volume of HClO4(aq) is [tex]V_{h} = 9.48 mL[/tex]
The concentration of HClO4(aq) is [tex]C_h = 0.200 M[/tex]
The Kb value is [tex]K_b = 6.5 * 10^{-5}[/tex]
Generally the the pOH of this reaction is mathematically represented as
[tex]pOH = pK_b + log [\frac{N_h}{N_b} ][/tex]
Here [tex]N_h[/tex] is the number of moles of acid which is evaluated as
[tex]N_h = C_h * V_h[/tex]
=> [tex]N_h = 0.200 * 9.48[/tex]
=> [tex]N_h = 1.896[/tex]
Here [tex]N_t[/tex] is the number of moles of acid which is evaluated as
[tex]N_t = C_t * V_t[/tex]
=> [tex]N_t = 0.100 * 20[/tex]
=> [tex]N_t = 2[/tex]
So
[tex]pOH = -log(K_b) + log [\frac{N_h}{N_b} ][/tex]
[tex]pOH = -log(6.5*10^{-5}) + log [\frac{1.896}{2} ][/tex]
=> [tex]pOH = 4.1639[/tex]
Generally the pH is mathematically represented as
[tex]pH = 14 - pOH[/tex]
=> [tex]pH = 14 - 4.1639[/tex]
=> [tex]pH = 9.84[/tex]
