# Question #c434b

Sep 27, 2016

Approx $0.2 \cdot m o l \cdot {L}^{-} 1$ with respect to ${H}_{2} S {O}_{4}$.

Given the stoichiometric equation, this is a $2 {C}_{2} {V}_{2} = {C}_{1} {V}_{1}$ equation.

#### Explanation:

As always we need a stoichiometrically balanced equation to show the molar equivalence.

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

As you already know, 2 equiv of lithium hydroxide react with 1 equiv sulfuric acid.

$\text{Moles of lithium hydroxide}$ $=$

$83.6 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.12 \cdot m o l \cdot {L}^{-} 1$

$=$ $0.0100 \cdot m o l$

And thus clearly, there were $\frac{0.0100 \cdot m o l}{2}$ ${H}_{2} S {O}_{4}$ in the initial $25 \cdot m L$ volume.

$\text{Concentration}$ $=$ $\text{Moles of sulfuric acid"/"Volume of solution}$

$=$ $\frac{5.02 \times {10}^{-} 3 \cdot m o l}{25 \times {10}^{-} 3 L}$ $=$ $0.201 \cdot m o l \cdot {L}^{-} 1$ with respect to ${H}_{2} S {O}_{4}$.

Given the stoichiometric equation, this is a $2 {C}_{2} {V}_{2} = {C}_{1} {V}_{1}$ equation.