# What is the concentration of sulfuric acid in the following scenario?

## A $10.0 \cdot m L$ volume of sulfuric acid, that had been prediluted TENFOLD, was titrated by a $67.02 \cdot m L$ volume of $N a O H \left(a q\right)$ of $6.000 \cdot m o l \cdot {L}^{-} 1$ concentration. What is $\left[{H}_{2} S {O}_{4}\right]$?

Jul 2, 2017

Unfeasibly high............

#### Explanation:

We perform the following acid base reaction.........

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

We add $67.02 \times {10}^{-} 3 L$ of sodium hydroxide with a concentration of $6.000 \cdot m o l \cdot {L}^{-} 1$......

$\text{Moles of NaOH} = 67.02 \times {10}^{-} 3 \cdot L \times 6.000 \cdot m o l \cdot {L}^{-} 1 = 0.40212 \cdot m o l$

And thus in the initial tritrand, the sulfuric acid solution, there were $0.20106 \cdot m o l$ sulfuric acid. Why did I HALVE the concentration with respect to $N a O H$?

But this solution had apparently been diluted TENFOLD..........

And thus there were $\frac{0.20106 \cdot m o l}{10.00 \times {10}^{-} 3 \cdot L} \ge 200 \cdot m o l \cdot {L}^{-} 1$ in the initial solution......This concentration is UNFEASIBLY HIGH. In the lab we can get 98% $\frac{w}{w}$, which is approx. $18 \cdot m o l \cdot {L}^{-} 1$ with respect to ${H}_{2} S {O}_{4}$..

So your question was chemically unreasonable, and the problem specification is WRONG.