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?

Apr 30, 2017

Approx. $10 \cdot m L$ of sulfuric acid need to be added.

Explanation:

By definition $p H + p O H = 14$ (in water under standard conditions).

And thus $p O H = 0.61$, and thus $\left[H {O}^{-}\right] = {10}^{- 0.61} \cdot m o l \cdot {L}^{-} 1.$

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

$\text{moles of}$ $H {O}^{-} = {10}^{- 0.61} \cdot m o l \cdot {L}^{-} 1 \times 47.5 \times {10}^{-} 3 \cdot L$

$= 1.17 \times {10}^{-} 2 \cdot m o l$

Given the stoichiometric equation.......

$2 N a O H \left(a q\right) + {H}_{2} S {O}_{4} \left(a q\right) \rightarrow N {a}_{2} S {O}_{4} \left(a q\right) + 2 {H}_{2} O \left(l\right)$

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

$\frac{1.17 \times {10}^{-} 2 \cdot m o l \times \frac{1}{2}}{0.602 \cdot m o l \cdot {L}^{-} 1} \times 1000 \cdot m L \cdot {L}^{-} 1 \cong 10 \cdot m L$