Question

What is the concentration of a `KOH (aq)` solution if 12.8 mL of this solution is required to react with 25.0 mL of .110 mol/L `H_2SO_4(aq)`?

Answer 1

We assume stoichiometric quantities of potassium hydroxide and sulfuric acid.

Explanation:

`2KOH(aq) + H_2SO_4 rarr K_2SO_4(aq) + 2H_2O(l)`

The reaction represents the neutralization of sulfuric acid by potassium hydroxide.

Moles of sulfuric acid: `25.0xx10^-3cancelLxx0.110*mol*cancel(L^-1)` `=` `??` `mol`

From the equation above, which shows that the stoichiometry of hydroxide to sulfuric acid is `2:1`, and the volume of stoichiometric acid:

`[KOH]=(25.0xx10^-3cancelLxx0.110*mol*cancel(L^-1)xx2)/(12.8xx10^-3L)` `=` `??mol*L^-1`.

We get an answer in `mol*L^-1`, as required.