An acidic solution is prepared by diluting a stock hydrochloric acid solution. If 12.76 mL of a 0.0576 M. HCl

Question

An acidic solution is prepared by diluting a stock hydrochloric acid solution. If 12.76 mL of a 0.0576 M. HCl(aq) sol. are diluted to a final vol. of 500 mL, what is the final pH of the solution? (assume complete dissociation of the strong acid)

Please explain how you got your work.. this problem has get me really stuck!

Answer 1

pH = -log(base 10) [H+]
so for complete dissociation
pH = -log(base 10) [acid]

concentration = number of moles / volume of solution
so the concentration of the diluted solution =
concentration of original solution * fraction of original acid solution in final solution
c = 0.0576 * (12.76 / 500)
c = 1.47 * 10 ^ -3

using this concentration, the pH can be calculated using
pH = -log(base 10) [acid]
where [acid] is concentration of acid = 1.47 * 10 ^ -3
so pH = -log(base 10) (1.47 * 10 ^ -3)
pH = 2.83

Answer 2

The number of moles must be the same, so the volume times the molarity for the concentrated solution will equal the volume times the molarity for the diluted solution. To get pH, you need to know the final molarity.

McVc = MdVd
Md = McVc/Vd = 0.0576*12.76/500 = 0.00147 M
pH = -log [H=] = -log (0.00147) = 2.83

Answer 3

According to dilution law of solution,MiVi=MfVf where,Mi=conc. of initial solution=11.6 M Vi=vol. of initial solution=15 ml Mf=conc. of final solution =? Vf=vol. of inal solution=500 ml Now,15 x 11.6=500 x Mf Mf=0.348 M (answer)