Question

What volume (in mL) of .158 M KOH solution is needed to titrate 20.0 mL of .0903 M H2SO4 to the equivalance point?

Answer 1

Approx. a `23*mL` volume of the sulfuric acid is required...........

Explanation:

We need (i), a stoichiometric equation.......

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

And (ii) equivalent quantities of `"sulfuric acid"`,

`"Moles of sulfuric acid"=20.0*cancel(mL)xx10^-3*cancelL*cancel(mL^-1)xx0.0903*mol*cancel(L^-1)=1.806xx10^-3*mol.`

And, CLEARLY, we need `"2 equiv potassium hydroxide......"`

`(2xx1.806xx10^-3*mol)/(0.158*mol*L^-1)xx10^3*mL*L^-1=22.9*mL.`

This is a good question because in the laboratory we would typically LIKE to deal with a titration whose TOTAL volume was around about `50*mL`.