Consider the weak acid ch3cooh (acetic acid). if a 0.048 m ch3cooh solution is 5.2% ionized, determine the [h3o+] concentration at equilibrium.

To determine the equilibrium concentration of hydronium ions in the solution, we use the given value of the percent ionized. Percent ionized is the percent of the ions that is dissociated into the solution. It is equal to the concentration of an ionized species over the initial concentration of the compound multiplied by 100 percent. For this case, the dissociation of the weak acid has a 1 is to 1 ratio to the ionized species such that the concentration of the CH3COO- and H+ ions at equilibrium would be equal. We calculate as follows:

5.2% = 5.2 M H3O+ / 100 M CH3COOH
5.2 M H3O+ / 100 M CH3COOH = [H3O+] / 0.048 M CH3COOH
[H3O+] = 0.2496 M 

Because electrons have negligible mass, to account for the mass of the isotope, there must be 16 neutrons, 16 neutrally charged, massive, fundamental particles present in the sulfur nucleus. and thus finally, if we have the 32s2− ion, there are 18 electrons, 16 protons, and 16 neutrons.

Answer is: the equilibrium  shifts toward the reactants, to the left side of chemical reaction.it is because lead(ii) nitrate dissociates in water: pb(no₃)₂ → pb²⁺(aq) + 2no₃⁻(aq).according to lechatelier's principle, the position of equilibrium moves to counteract thechange (more lead cations, common ion), the position of equilibrium will move so that the concentration ofproducts decrease and concentration of reactants of chemical reaction increase.read more on -