Respuesta :
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 [tex]$HC_6H_5CO_2$[/tex] with a 0.3600 M solution of KOH. The [tex]pK_a[/tex] 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 [tex]$C_6H_5OCOOH$[/tex] [tex]$=148.9 \ mL \times \frac{L}{1000\ mL} \times \frac{1.100 \ mol}{L}$[/tex]
= 0.16379 mol
Number of moles of NaOH added [tex]$=232.0 \ mL \times \frac{L}{1000\ mL} \times \frac{0.3600 \ mol}{L}$[/tex]
= 0.08352 mol
ICE table :
[tex]C_6H_5OCOOH \ \ \ + \ \ \ OH^- \ \ \ \rightarrow \ \ C_6H_5OCOO^- \ \ \ \ + \ \ H_2O[/tex]
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 [tex]$C_6H_5OCOOH, [C_6H_5OCOOH]$[/tex] [tex]$=\frac{0.08027 \ mol}{0.3809 \ L}$[/tex]
= 0.211 M
Concentration of [tex]$C_6H_5OCOO^- , [C_6H_5OCOO^-] =\frac{0.08352 \ mol}{0.3809 \ L}[/tex]
= 0.219 M
[tex]pK_a[/tex] of [tex]C_6H_5OCOOH = 4.20[/tex]
According to Henderson equation,
[tex]$pH = pK_a + \log \frac{[C_6H_5OCOO^-]}{[C_6H_5OCOOH]}[/tex]
[tex]$=4.20 + \log \frac{0.219}{0.211}$[/tex]
= 4.22
Therefore, the pH of the acid solution is 4.22
