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.

Answer 1

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
`pH = 9.84`

Step-by-step explanation:

From the question we are told that

The volume of trimethylamine, (CH3)3N(aq) is
`V_(t) = 20.00mL`

The concentration of trimethylamine is
`C_t = 0.1000 \ M`

The volume of HClO4(aq) is
`V_(h) = 9.48 mL`

The concentration of HClO4(aq) is
`C_h = 0.200 M`

The Kb value is
`K_b = 6.5 * 10^(-5)`

Generally the the pOH of this reaction is mathematically represented as

`pOH = pK_b + log [(N_h)/(N_b) ]`

Here
`N_h` is the number of moles of acid which is evaluated as

`N_h = C_h * V_h`

=>
`N_h = 0.200 * 9.48`

=>
`N_h = 1.896`

Here
`N_t` is the number of moles of acid which is evaluated as

`N_t = C_t * V_t`

=>
`N_t = 0.100 * 20`

=>
`N_t = 2`

So

`pOH = -log(K_b) + log [(N_h)/(N_b) ]`

`pOH = -log(6.5*10^(-5)) + log [(1.896)/(2) ]`

=>
`pOH = 4.1639`

Generally the pH is mathematically represented as

`pH = 14 - pOH`

=>
`pH = 14 - 4.1639`

=>
`pH = 9.84`