0.035 miles of Acetic Acid (CH3COOH) are dissolved in 150mL of water. What is the pH of this solution? Ka=1.58x10^-5

Answer 1

To find the pH of the solution, we first need to calculate the concentration of acetic acid in the solution.

First, convert 0.035 miles to moles:
0.035 moles x (1 mile/1,609.34 meters) x (1000 mL/1 meter^3) x (1 L/1000 mL) = 2.17 x 10^-4 moles

Then, calculate the molarity of acetic acid in the solution:
Molarity = moles/volume
Molarity = 2.17 x 10^-4 moles / 0.15 L = 0.00145 M

Since acetic acid is a weak acid, it will partially ionize in water:
CH3COOH ⇄ CH3COO- + H+

Now, we can use the equation for the ionization of weak acids to find the concentration of H+ ions:
Ka = [CH3COO-][H+] / [CH3COOH]
1.58x10^-5 = (x)(x) / (0.00145-x)
x = [H+] = 0.00129 M

Therefore, the pH of the solution is:
pH = -log[H+]
pH = -log(0.00129)
pH = 2.89

Therefore, the pH of the solution is 2.89.