Question

Hello,

Could you please provide a detail explanation as to how to find each of these steps.

The ionization constant of a very weak acid, HA, is 5.2×10^-9. Calculate the equilibrium concentrations of H₃O⁺, A^-, and HA in a 0.040 M solution of the acid.

Determine the concentrations of all species at equilibrium and the solution pH.

[H₃O⁺] =  mol/L

[A^-] =  mol/L

[HA] =  mol/L

pH =

Answer 1

Ka for HA= 5.2 x 10 9 H3o+ НА + H2O оочом + 4- нро - Х (ооцо-х) - л + ka- [Hot] [RO] THAJ xxx = 512 X10-9 oo Чо- х? = зох)о-1° — 52х0-3х х? + 5.2 xio 9x -.os xo-= x = -5.2х10 9 + (52xio-9)2 + х = oxid' 2 . = -5.2 х10 9 + q qч хоs Фо = 1.992 x 10 -5 м [H₃O+7= 1.442x10-5 M. свој = 1.9ч хо 5 м [HAIF 0.040 – 1.442x105 = 3gyax 102 м pнa - Toy ( 1 уу? xo 5) pH = Чу