Question

Calculate the pH of a 0.095 M weak acid solution that is 1.9% ionized.

		a. 3.74
		b. 0.744
		c. 2.74
		d. 1.72
		e. 1.02

Answer 1

Let represent the dissociation of weak acid as,

HA ---- > H⁺ + H^-

Percent ionization is defines as

% ionization = ( [H⁺])/ [HA) ) *100           .............(1)

We are given 0.095 M weak acid , hence [HA] = 0.095 M

percent ionization is given as 1.9 %

Let us substitute it in equation (1) to obtain the value of [H⁺]

1.9 =( ([H⁺])/0.095 )*100

[H⁺]    =(1.9*0.095)/100 = 1.805*10^-3 M

Now concentration value can be converted to pH value by the following relation

pH = - Log([H⁺] )

pH = - Log (1.805*10^-3) = 2.74

Hence correct answer is option (C) 2.74