Question

Phenol, commonly called carbolic acid, is a weak organic acid. If you dissolve 0.892 g of the acid in enough water to make 357 mL of solution, what is the equilibrium hydronium ion concentration? And what is the pH o the solution?

Answer 1

First we must know the concentration of phenol solution.

We have, No. of moles = Mass / Molar mass

$\therefore$ No. of moles of phenol = 0.892 g / 94.11 g/mol = 0.009478 mol

Volume of solution = 357 ml = 0.357 L

We have, Molarity = No. of moles of solute / volume of solution in L

[C₆H₅OH] = 0.009478 mol / 0.357 L = 0.02655 M

Consider dissociation of phenol in water.

C₆H₅OH (aq) + H₂O $\rightleftharpoons$ C₆H₅O ^- (aq) + H₃O ⁺ (aq)

Equilibrium constant expression for above reaction is K a = [H₃O ⁺] [C₆H₅O ^- ] / [C₆H₅OH] = 1.01 $\times$ 10 ^-10

Let's use ICE table to find out equilibrium concentrations of C₆H₅O ^- , H₃O ⁺ and C₆H₅OH .

Concentration (M)	C6H5OH	H3O +	C6H5O -
Initial	0.02655
Change	-X	+X	+X
Equilibrium	0.02655 -X	X	X

Putting above values in Ka formula we get

Ka = (X) (X) / 0.02655 - X = 1.01 $\times$ 10 ^-10

Phenol is a weak acid, hence we can assume X is very small as compared 0.02655. So we can write 0.02655 -X $\simeq$ 0.02655.

Therefore, K a = X ² / 0.02655 = 1.01 $\times$ 10 ^-10

X ² = 1.01 $\times$ 10 ^-10 (0.02655)

X ² = 2.682 $\times$ 10 ^-12

X = 1.638  $\times$ 10 ^-06 M = [H₃O ⁺] = [C₆H₅O ^- ]

We have, pH = -log [H₃O ⁺] = - log 1.638  $\times$ 10 ^-06 = 5.78

ANSWER : Equilibrium hydronium ion concentration = 1.64  $\times$ 10 ^-06 M

pH of phenol solution = 5.78