Question

ph of 0.001 M acetic acid would be

Answer 1

Your answer is 3.87.
Step by step explanation

CH3COOH——————→ CH3COO- + H+

initial= 0.001(ch3cooh), 0(ch3coo- and h+ (no disssociation yet))

Final (after dissociation)= 0.001*(1-a), ca( h+ and ch3coo-), where a is degree of dissociation and c= 0.001

now according to equilibrium law Ka = concentration of production/concen. of reactants

acetic acid dissociates ( partially, couldn’t show the equilibrium arrows) to give H+ and acetate ion.

now Ka value of ch3cooh= 1.75 * 10^-5

1.76 * 10^-5 = (ca * ca)/ c(1-a) =( approx. ) ca^2/ 1

1.76 * 10^-5 = 0.001 a^2

; a= 1.32 *10^-1 = 0.132

; concentration of h+ ions = ca= 0.001* 0.132 = 1.32 *10^-4

now pH = -log(H+)

pH= -log(1.34 * 10^-4)

pH= -(0.1271 -4)

pH = 3.87