Question

If a 1.00 ml sample of the reaction mixture for the equilibrium constant experiment required 32.40 ml of 0.258 m naoh to titrate it, what is the monoprotic acetic acid concentration in the mixture?

Answer 1

The concentration of the acetic acid is  8.36M

For given question,

Volume of acetic acid = 1.00 mL

                                     = 0.001 L

Volume of base(NaOH) = 32.40  mL

                                       = 0.0324L

Concentration of the base = 0.258 M

First consider the neutralization reaction between sodium hydroxide and acetic acid.

CH₃COOH + NaOH → CH₃COONa + H₂O

We need to obtain the unknown concentration of the acid used in the reaction.

The number of moles of the base used in the reaction using the equation below:

Number of moles of base = concentration of base x volume

⇒ Number of moles of base = 0.258 × 0.0324  

⇒ Number of moles of base = 0.0084 mol

From the balanced reaction equation, we know that:

1mole of the base neutralize on mole of acid,

Therefore, 0.0084 mol of the acid would be neutralized by the base.

To find the concentration of the acetic acid,

Concentration of acid = (Number of moles of acid) / (volume of the acid)

⇒ Concentration of acid = 0.0084 / 0.001

⇒ Concentration of acid = 8.36M

Therefore,  the concentration of the acetic acid is  8.36M

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