Question

In a titration experiment 43.18 mL of 0.283 M KOH is reacted with H2SO4. The endpoint is reached when 44.62 mL of the acid is mixed with the base. What is the molar concentration of the acid?

Answer 1

The balanced formula of the reaction described is:

`2KOH_((aq))+H_2SO_(4(aq))→K_2SO_(4(aq))+2H_2O_((l))`

To find the molar concentration (Molarity) of the solution we will follow the following steps:

1. We find the moles of KOH present in the basic solution using the molarity equation that tells us:

`Molarity=(MolesSolute)/(Lsolution)`
`\begin{gathered} MolesSolute=Molarity* Lsolution \\ MolesSolute=0.283M*0.04318L \\ MolesSolute=0.012molKOH \end{gathered}`

2. From the stoichiometry of the reaction we find the moles of H2SO4 needed to neutralize the moles of KOH. The ratio H2SO4 to KOH is 1/2.

`\begin{gathered} molH_2SO_4=givenmolKOH*(1molH_2SO_4)/(2molKOH) \\ molH_2SO_4=0.012molKOH*(1molH_(2)SO_(4))/(2molKOH)=0.006molH_2SO_4 \end{gathered}`

3. We find the molarity of the solution using the molarity formulation from point 1.

`\begin{gathered} Molarity=(MolesSolute)/(Lsolution) \\ Molarity=(0.006molH_2SO_4)/(0.04462L)=0.137M \end{gathered}`

Answer: The molarity or molar concentration of the acid solution is 0.137M