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

Question

8075 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-03T17:42:37

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

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 $50*mL$ .