# 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?

Dec 14, 2016

The $\text{pH}$ will be 2.08.

#### Explanation:

The strong acid $\text{HCl}$ will almost completely suppress the ionization of the weak acid $\text{HAc}$.

Thus, we need to consider only the contribution of $\text{H"_3"O"^"+}$ from the $\text{HCl}$.

The equation for the dissociation of $\text{HCl}$ is

$\text{HCl + H"_2"O" → "H"_3"O"^"+" + "Cl"^"-}$

$\text{Moles of HCl" = 0.0100 color(red)(cancel(color(black)("L HCl"))) × "0.050 mol HCl"/(1 color(red)(cancel(color(black)("L HCl")))) = "0.000 50 mol HCl}$

Since $\text{HCl}$ is a strong acid, it will dissociate completely to form 0.0050 mol of $\text{H"_3"O"^"+}$.

The volume of the solution is

$V = \text{10.0 mL + 50.0 mL" = "60.0 mL" = "0.060 L}$

["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