Question

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?

Answer 1

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`