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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.

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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

User Douglas Ribeiro
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