Question

If 25.98 mL of a standard 0.1180 M KOH solution reacts with 52.50 mL of solution, what is the molarity of the acid solution?

Short answer

The molarity of the acid solution is 0.058 M.

Step-by-step solution

Write the balanced molecular equation

The reaction of potassium hydroxide and acetic acid is shown below.

$KOH\left(aq\right)+C{H}_{3}COOH\left(aq\right)\to C{H}_{3}COOK\left(aq\right)+H_{2}O\left(l\right)$

Calculate the moles of acetic acid

Given values are:

$$\begin{gathered} volumeofstandard0.1180MKOHsolution=25.98ml \\ volumeofaceticacidsolution=52.50ml \end{gathered}$$

Moles of acetic acid is:

$\begin{array}{rcl}molesofaceticacid& =& 25.98mlKOH\times \frac{1lit}{1000ml}\times \frac{0.1180molKOH}{1litKOHsolution}\times \frac{1molC{H}_{3}COOH}{1molKOH}\\ & =& 3.07x{10}^{-3}mol C{H}_{3}COOH\end{array}$

Calculate the molarity of acetic acid

The molarity of the acid solution is:

$molarityofacidsolution=\frac{molesofaceticacid}{volumeofaceticacid}$

The molarity of acetic acid is calculated as:

$$\begin{gathered} =\frac{3.07x{10}^{-3}molC{H}_{3}COOH}{(52.50ml)\left(\frac{1lit}{1000ml}\right)} \\ =0.058MC{H}_{3}COOHsolution \end{gathered}$$