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

1 Answer
Sep 19, 2017

Answer:

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

Explanation:

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.....