# 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)?

Sep 19, 2017

We need a $2.0 \cdot m L$ volume.......

#### Explanation:

We need (i) a 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 (ii) we need equivalent quantities of each reagent....

$\text{Moles of NaOH} = 20 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.1 \cdot m o l \cdot {L}^{-} 1 = 2 \times {10}^{-} 3 \cdot m o l .$

And we thus need a $1 \times {10}^{-} 3 \cdot m o l$ quantity with respect to the DIACID ${H}_{2} S {O}_{4}$....

$V = \frac{n}{C} = \frac{1 \times {10}^{-} 3 \cdot m o l}{0.5 \cdot m o l \cdot {L}^{-} 1} \times {10}^{3} \cdot m L \cdot {L}^{-} 1 = 2.0 \cdot m L$

Clearly, we should titrate the base with a LESS concentrated acid.....