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?

1 Answer
Apr 30, 2017

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

Explanation:

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