What is the pH of a 0.11 M solution of C6H5OH (Ka = 1.3 x 10-10)?

The pH is 5.42

Explanation:

Given the following;

Ka = 1.3*10^-10 pH =? 0.11m C6H5OH

We would solve the question by using the I.C.E table,

where; I represents Initial Concentration

C represents Change in Concentration

E represents Equilibrium

I 0.11m 0 0

C -x +x +x

E 0.11-x x x

The formula, Ka = ([H^{+}] [A^{-}])/[HA]

Ka = ([H+] [C6H5O-])/[C6H5OH]

substituting into the above equation gives;

1.3*10^-10 = [x.x]/[0.11-x]

1.3*10^-10 = x^2/0.11-x

Assume x << 0.11 because C6H5OH dissociation can be neglected. Hence, (0.11-x) approximately equals to 0.11

Therefore,

pH = -log(H+)

pH = -log(3.78153*10^-6)

pH = -(-5.42)

pH = 5.42

it would be d. henry moseley

hope this !

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