What is the volume of 0.602 M H2SO4 necessary to neutralise completely the hydroxide in 47.5 mL of NaOH having pH 13.39?

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Answer 1 · 2017-04-30T07:02:03

Approx. #10*mL# of sulfuric acid need to be added.

By definition #pH+pOH=14# (in water under standard conditions).

And thus #pOH=0.61#, and thus #[HO^(-)]=10^(-0.61)*mol*L^-1.#

And thus..........

#"moles of"# #HO^(-)=10^(-0.61)*mol*L^-1xx47.5xx10^-3*L#

#=1.17xx10^-2*mol#

Given the stoichiometric equation.......

#2NaOH(aq)+H_2SO_4(aq)rarrNa_2SO_4(aq) + 2H_2O(l)#

And we need to add a volume of.........

#(1.17xx10^-2*molxx1/2)/(0.602*mol*L^-1)xx1000*mL*L^-1~=10*mL#