# 29.4mL of an CH_3COOH solution were titrated with 18.5mL of a 0.175M LiOH solution to reach the equivalence point What is the molarity of the CH_3COOH solution?

Jul 4, 2016

$\textsf{0.11 \textcolor{w h i t e}{x} \text{mol/l}}$

#### Explanation:

sf(CH_3COOH_((aq))+LiOH_((aq))rarrCH_3COO^(-)Li_((aq))^(+)+H_2O_((l))

Concentration = amount of solute / volume of solution

$\textsf{c = \frac{n}{v}}$

$\therefore$$\textsf{n = c \times v}$

$\therefore$$\textsf{n L i O H = 0.175 \times \frac{18.5}{1000} = 0.003238}$

Where $\textsf{n}$ refers to the number of moles.

From the 1:1 molar ratio as shown by the equation, we can say that the number of moles of ethanoic acid must be the same.

$\therefore$$\textsf{\left[C {H}_{3} C O O {H}_{\left(a q\right)}\right] = \frac{n}{v} = \frac{0.003238}{\frac{29.4}{1000}} = 0.11 \textcolor{w h i t e}{x} \text{mol/l}}$

Note that I have converted $\textsf{m l}$ to $\textsf{l i t r e}$ by dividing by 1000.