Problem: 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.

FREE Expert Solution

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:


Recall:

Molarity(M)=moles solute (mol)liters solution (L)

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


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Problem Details

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.

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