Question

Calculate the pH of a given solution

0.1M HAc, pKA=4.76

HAc + H2O = H3O+ + Ac-

Answer 1

Ans -

Chemical Equation we have -

HAc = CH₃COOH

Ac^- = CH₃COO^-

CH₃COOH + H₂O ----------> H3O⁺ + CH₃COO^-

Ka = Dissociation constant

-pKA = log Ka

Ka = 10^-4.76

Ka = 1.74 x10 ^-5

Ka = Product concentration / reactant concentration

Ka =  [H3O⁺] [CH₃COO^-] / CH₃COOH

let concentration after dissociation be -

CH₃COOH + H₂O ----------> H3O⁺ + CH₃COO^-

(0.1-x) (x) (x)

Putiing the values in dissociation constant Ka

1.74 x10^-5 = (x ). (x) / (0.1-x)

1.74 x10^-5 = (x )² / (0.1-x)

neglecting x in denominator as it very small than 0.1 we have -

1.74 x10^-5 = (x )² / (0.1)

x = (1.74 x10^-5 x 0.1) ^0.5

X = 1.32x10^-3 or 0.00132

[H3O⁺] and [CH₃COO^-] = 0.00132 M

we know that pH = -log [H3O⁺]

pH = -log [.00132]

pH =2.879