Question

Acid - Base TitrationWhat is the molarity of an acetic acid solution that required 22.5 mL of a 0.185 M NaOH solution to neutralize 15.0 mL of the acid solution? Write the balanced equation for the reaction,

Answer 1

Answer: the molarity of the acetic acid solution is 0.278 M

Explanation:

The question requires us to determine the molarity of an acetic acid (CH3COOH) solution, given the data for a titration of this solution with sodium hydroxide (NaOH).

The following information was provided by the question:

Molarity of NaOH = M(OH-) = 0.185 M

Volume of NaOH = V(OH-) = 22.5 mL

Volume of CH3COH = V(H+) = 15.0 mL

The neutralization reaction between NaOH and CH3COOH produces the salt sodium acetate (NaCH3COO) and water (H2O), and its balanced chemical equation can be written as:

$CH_3COOH_{(aq)}+NaOH_{(aq)}\rightarrow NaCH_3COO_{(aq)}+H_2O_{(l)}$

At the neutralization point, the amount of base (OH- ions) and acid (H+ ions) must be the same. Thus, we can say that when the acid is neutralized:

$n_{H^+}=n_{OH^-}$

where n is the number of moles.

Considering the definition of molarity, where the number of moles (n, in mol) is divided by the volume of the sample (V, in liters), we can write an expression for the number of moles:

$M=\frac{n}{V}\rightarrow n=M\times V$

And we can rewrite the expression for the neutralization point as:

$n_{H^+}=n_{OH^-}\rightarrow M_{H^+}\times V_{H^+}=M_{OH^-}\times V_{OH^-}$

Using the expression above and applying the values provided by the question, we can determine the molarity of the acetic acid solution:

$\begin{gathered} M_{H^+}\times V_{H^+}=M_{OH^-}\times V_{OH^-}\rightarrow M_{H^+}=\frac{M_{OH^-}\times V_{OH^-}}{V_{H^+}} \\ M_{H^+}=\frac{0.185M\times22.5mL}{15.0mL}=0.278M \end{gathered}$

Therefore, the molarity of the acetic acid solution is 0.278 M.