Question

Calculate the pH of a solution formed by the addition of 10.0mL of 0.050M hydrochloric acid to a 50.0mL sample of 0.20M acetic acid?

Consider the dissociation of acetic acid
HAc <-> Ac- + H+(aq) Ka=1.8x10-5.

Calculate the pH of a solution formed by the addition of 10.0mL of 0.050M
hydrochloric acid to a 50.0mL sample of 0.20M acetic acid?

Answer 1

The `"pH"` will be 2.08.

Explanation:

The strong acid `"HCl"` will almost completely suppress the ionization of the weak acid `"HAc"`.

Thus, we need to consider only the contribution of `"H"_3"O"^"+"` from the `"HCl"`.

The equation for the dissociation of `"HCl"` is

`"HCl + H"_2"O" → "H"_3"O"^"+" + "Cl"^"-"`

`"Moles of HCl" = 0.0100 color(red)(cancel(color(black)("L HCl"))) × "0.050 mol HCl"/(1 color(red)(cancel(color(black)("L HCl")))) = "0.000 50 mol HCl"`

Since `"HCl"` is a strong acid, it will dissociate completely to form 0.0050 mol of `"H"_3"O"^"+"`.

The volume of the solution is

`V= "10.0 mL + 50.0 mL" = "60.0 mL" = "0.060 L"`

`["H"_3"O"^"+"] = "moles"/"litres" = "0.000 50 mol"/"0.060 L" = "0.008 33 mol/L"`

`"pH" = -log["H"_3"O"^"+"] = "-"log("0.00 833") = 2.08`