The substance benzoic acid (C6H5COOH) is a weak acid (Ka = 6.3×10^-5).

The substance benzoic acid (C6H5COOH) is a weak acid (Ka = 6.3×10^-5). What is the pH of a 5.09×10^-2 M aqueous solution of potassium benzoate, KC6H5COO

The Correct Answer and Explanation is:

Answer: The pH of a 

5.09×10−25.09×10−2

 M aqueous solution of potassium benzoate is 8.45.

Explanation

This problem requires the determination of the pH of a salt solution. The salt in question, potassium benzoate (

KC6H5COOKC6​H5​COO

), is formed from a strong base (potassium hydroxide, KOH) and a weak acid (benzoic acid, 

C6H5COOHC6​H5​COOH

). When dissolved in water, the salt dissociates completely into its constituent ions:

KC6H5COO(aq)→K+(aq)+C6H5COO−(aq)KC6​H5​COO(aq)→K+(aq)+C6​H5​COO−(aq)

The potassium ion (

K+K+

) is the conjugate acid of a strong base and is thus a spectator ion that does not react with water. However, the benzoate ion (

C6H5COO−C6​H5​COO−

) is the conjugate base of a weak acid. It will react with water in a hydrolysis reaction, accepting a proton and producing hydroxide ions (

OH−OH−

), making the solution basic.

C6H5COO−(aq)+H2O(l)⇌C6H5COOH(aq)+OH−(aq)C6​H5​COO−(aq)+H2​O(l)⇌C6​H5​COOH(aq)+OH−(aq)

To calculate the concentration of 

OH−OH−

 at equilibrium, the base dissociation constant (

KbKb​

) for the benzoate ion is needed. It can be calculated from the acid dissociation constant (

KaKa​

) of its conjugate acid, benzoic acid, using the ion-product constant for water, 

Kw=1.0×10−14Kw​=1.0×10−14

.

Kb=KwKa=1.0×10−146.3×10−5=1.59×10−10Kb​=Ka​Kw​​=6.3×10−51.0×10−14​=1.59×10−10

An ICE table can be used to find the equilibrium concentration of 

OH−OH−

. The initial concentration of the benzoate ion is 

5.09×10−25.09×10−2

 M.

	C6H5COO−C6​H5​COO−	C6H5COOHC6​H5​COOH	OH−OH−
Initial	5.09×10−25.09×10−2	0	0
Change	-x	+x	+x
Equilibrium	(5.09×10−2−x)(5.09×10−2−x)	x	x

The 

KbKb​

 expression is:

Kb=[C6H5COOH][OH−][C6H5COO−]=(x)(x)(5.09×10−2−x)Kb​=[C6​H5​COO−][C6​H5​COOH][OH−]​=(5.09×10−2−x)(x)(x)​

Since 

KbKb​

 is very small, we can assume 

xx

 is negligible compared to the initial concentration, simplifying the equation to:

1.59×10−10≈x25.09×10−21.59×10−10≈5.09×10−2x2​

x2=(1.59×10−10)(5.09×10−2)=8.09×10−12x2=(1.59×10−10)(5.09×10−2)=8.09×10−12

x=[OH−]=8.09×10−12=2.85×10−6 Mx=[OH−]=8.09×10−12​=2.85×10−6 M

Now, the pOH can be calculated, followed by the pH.

pOH=−log⁡[OH−]=−log⁡(2.85×10−6)=5.55pOH=−log[OH−]=−log(2.85×10−6)=5.55

pH=14.00−pOH=14.00−5.55=8.45pH=14.00−pOH=14.00−5.55=8.45