Question

An analytical chemist is titrating of a solution of benzoic acid with a solution of . The of benzoic acid is . Calculate the pH of the acid solution after the chemist has added of the solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of solution added.

Answer 1

The question is incomplete. The complete question is :

An analytical chemist is titrating 148.9 mL of a 1.100 M solution of benzoic acid
`$HC_6H_5CO_2$` with a 0.3600 M solution of KOH. The
`pK_a` of benzoic acid is 4.20. Calculate the pH of the acid solution after the chemist has added 232.0 mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added.

Solution :

Number of moles of
`$C_6H_5OCOOH$`
`$=148.9 \ mL * (L)/(1000\ mL) * (1.100 \ mol)/(L)$`

= 0.16379 mol

Number of moles of NaOH added
`$=232.0 \ mL * (L)/(1000\ mL) * (0.3600 \ mol)/(L)$`

= 0.08352 mol

ICE table :

`C_6H_5OCOOH \ \ \ + \ \ \ OH^- \ \ \ \rightarrow \ \ C_6H_5OCOO^- \ \ \ \ + \ \ H_2O`

I (mol) 0.16379 0.08352 0

C (mol) -0.08352 -0.08352 +0.08352

E (mol) 0.08027 0 0.08352

Total volume = (148.9 + 232) mL

= 380.9 mL

= 0.3809 L

Concentration of
`$C_6H_5OCOOH, [C_6H_5OCOOH]$`
`$=(0.08027 \ mol)/(0.3809 \ L)$`

= 0.211 M

Concentration of
`$C_6H_5OCOO^- , [C_6H_5OCOO^-] =(0.08352 \ mol)/(0.3809 \ L)`

= 0.219 M

`pK_a` of
`C_6H_5OCOOH = 4.20`

According to Henderson equation,

`$pH = pK_a + \log ([C_6H_5OCOO^-])/([C_6H_5OCOOH])`

`$=4.20 + \log (0.219)/(0.211)$`

= 4.22

Therefore, the pH of the acid solution is 4.22