We are asked to solve for the **pH of acetic acid ****dissolved in water**.

Since HC_{2}H_{3}O_{2} has a low K_{a} value, it’s a weak acid. Remember that ** weak acids** partially dissociate in water and that

**HC _{2}H_{3}O_{2}_{(aq)} + H_{2}O_{(l)} **

**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 [H_{3}O^{+}].

*Step 4*: Calculate the pH.

**Step1**: We need to get the **molarity of the acetic acid solution**** **first**:**

Recall:

$\overline{){\mathbf{Molarity}}{\mathbf{\left(}}{\mathbf{M}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{moles}\mathbf{}\mathbf{solute}\mathbf{}\mathbf{\left(}\mathbf{mol}\mathbf{\right)}}{\mathbf{liters}\mathbf{}\mathbf{solution}\mathbf{}\mathbf{\left(}\mathbf{L}\mathbf{\right)}}}$

*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 HC _{2}H_{3}O_{2} is:**

**HC _{2}H_{3}O_{2 }**

** 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 HC_{2} H_{3} O_{2}) 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.