Consider a 0.10 m aqueous benzoic acid, cehecooh. the k benzoic acid. 6.5 x 10 for a) write a balanced equation that shows the reaction of benzoic acid with water and construct an appropriate i. c.e. table. [3 pts] b) calculate the [hao'] at equilibrium. [4 pts] c) calculate the ph of this solution. [1 pt

a) C6H5COOH + H2O ↔ H3O+  +  C6H5COO-

b) [ H3O+ ] = 2.517 E-3 M

c) pH = 2.599

Explanation:

a) balanced equation:

C6H5COOH + H2O ↔ H3O+  +  C6H5COO-

⇒ Ka = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ] = 6.5 E-5

mass balance:

0.10 m = [ C6H5COO- ] + [ C6H5COOH ](1)

charge balance:

[ H3O+ ] = [ C6H5COO- ] + [ OH- ] [ OH- ] : comes from water, it's not significant

⇒ [ H3O+ ] = [ C6H5COO- ] (2)

b) (2) in (1):

⇒ 0.10 M = [ H3O+ ] + [ C6H5COOH ]

⇒ [ C6H5COOH ] = 0.10 - [ H3O+ ]

⇒ Ka = [ H3O+ ]² / ( 0.1 - [ H3O+ ] ) = 6.5 E-5

⇒ [ H3O+ ]² + 6.5 E-5 [ H3O+ ] - 6.5 E-6 = 0

⇒ [ H3O+ ] = 2.517 E-3 M

c) pH = - log [ H3O+ ]

⇒ pH = - Log ( 2.517 E-3 )

⇒ pH = 2.599