We are asked to solve for the pH of acetic acid dissolved in water.
Since HC2H3O2 has a low Ka value, it’s a weak acid. Remember that weak acids partially dissociate in water and that acids donate H+ to the base (water in this case). The dissociation of HC2H3O2 is as follows:
HC2H3O2(aq) + H2O(l) ⇌ H3O+(aq) + HC2H3O2 –(aq); Ka = 1.8 × 10–5 (found on books and on the internet)
We shall follow these steps in solving the problem:
Step 1: Get the molarity of the acetic acid solution.
Step 2: Use ICE table to determine the equilibrium concentrations (i.e. the expressions).
Step 3: Determine [H3O+].
Step 4: Calculate the pH.
Step1: We need to get the molarity of the acetic acid solution first:
initial volume(× density)→mass of acetic acid (÷ molar mass of acetic acid)→moles of acetic acid(÷ volume of the final solution, L)→molarity of the solution
density of glacial acetic acid = 1.05 g/mL
volume of final solution=1.50 L
The molar mass of HC2H3O2 is:
HC2H3O2 2 C × 12.01 g/mol C = 24.02 g/mol
4 H × 1.008 g/mol H = 4.032 g/mol
2 O × 16.00 g/mol O = 32.00 g/mol
Sum = 60.052 g/mol
If 20.0 mL of glacial acetic acid (pure HC2 H3 O2) is diluted to 1.50 L with water, what is the pH of the resulting solution? The density of glacial acetic acid is 1.05 g / mL.
Frequently Asked Questions
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Weak Acids concept. If you need more Weak Acids practice, you can also practice Weak Acids practice problems.