please help with this question

Question

Acetic (ethanoic) acid is the most common weak acid used in industry.  Determine the pH and pOH of an acetic acid solution prepared by dissolving 60.0 kg of pure, liquid acetic acid, to make 1.25 kL of solution.  The percent reaction with water at this concentration is 0.48%. Show all your work.

Answer 1

To solve this problem first you need to find the concentration (M) of the weak acid.

M = n / l

In the equation above n is the number of moles and l is the amount of liters.

The number of moles is equal to the mass in grams divided by the molar mass.

So n = g / MM

Converting 60 kg to g is 60000g and the molar mass is 60.

Therefore the number of moles is 60000 / 60 or 1000 moles.

To find the concentration you need to divide the moles by the amount in liters.

M = 1000 / 1250 = 0.8 M

Now that you know the concentration we need to determine how much of that concentration dissociates. The question says that 0.48% of it will break apart. Therefore we can use the following to determine how much dissociates.

0.48% = x / 0.8

Solving for x gives 0.00384 M.

This means that after the weak acid broke apart we would now have 0.00384 of each ion (including the hydrogen ion) in solution.

To find the pH you use the equation pH = -log [ H+ ]

So the pH = - log 0.00384 = 2.41

To find the pOH you can use the following equation 14 = pH + pOH to get 11.58.