What volume of $0.5molL^-1$ $"sulfuric acid"$ is required for equivalence with a $20mL$ volume of $0.5mol*L^-1$ $NaOH(aq)$ ?

Question

1849 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-19T15:47:32

We need a $2.0*mL$ volume.......

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

$2NaOH(aq) + H_2SO_4(aq) rarr Na_2SO_4(aq) + 2H_2O(l)$

And (ii) we need equivalent quantities of each reagent....

$"Moles of NaOH"=20*mLxx10^-3*L*mL^-1xx0.1*mol*L^-1=2xx10^-3*mol.$

And we thus need a $1xx10^-3*mol$ quantity with respect to the DIACID $H_2SO_4$ ....

$V=n/C=(1xx10^-3*mol)/(0.5*mol*L^-1)xx10^3*mL*L^-1=2.0*mL$

Clearly, we should titrate the base with a LESS concentrated acid.....

What volume of 0.5*mol*L^-10.5*mol*L^-1 "sulfuric acid""sulfuric acid" is required for equivalence with a 20*mL20*mL volume of 0.5*mol*L^-10.5*mol*L^-1 NaOH(aq)NaOH(aq)?