Question

If 28.7mL of 0.205M KOH is required to completely neutralize 20.0mL of a HC2H3O2 solution, what is the molarity of the acetic acid solution?
HC2H3O2(aq) + KOH(aq) → KC2H3O2(aq) + H2O(l)

Answer 1

Answer on Question #49742 - Chemistry - Other

Question

If 28.7mL of 0.205M KOH is required to completely neutralize 20.0mL of a HC2H3O2 solution, what is the molarity of the acetic acid solution?

$$\mathrm{HC2H3O2(aq)} + \mathrm{KOH(aq)} \rightarrow \mathrm{KC2H3O2(aq)} + \mathrm{H_2O(l)}$$

Answer:

According to the reaction, 1 mole of acetic acid reacts with 1 mole of KOH, therefore, if $\mathrm{HC_2H_3O_2}$ is completely neutralized, there are no moles of $\mathrm{HC_2H_3O_2}$ in solution, so number of moles of KOH required to completely neutralize acetic acid is equal to initial number of moles of the acid:

$$n(\mathrm{KOH}) = n(\mathrm{HC_2H_3O_2})$$

Number of moles of substance in solution equals:

$$n = CV$$ $$C(\mathrm{KOH}) \cdot V(\mathrm{KOH}) = C(\mathrm{HC_2H_3O_2}) \cdot V(\mathrm{HC_2H_3O_2})$$

Molar concentration or molarity of acetic acid solution equals:

$$C(\mathrm{HC_2H_3O_2}) = C(\mathrm{KOH}) \cdot V(\mathrm{KOH}) / V(\mathrm{HC_2H_3O_2}) = 0.205 \cdot 28.7 / 20.0 = 0.294\ \mathrm{M}$$

Answer: 0.294 M

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