# A 38.74*mL volume of 0.50*mol*L^-1 sodium hydroxide solution reaches a stoichiometric endpoint with 19.37xx10^-3*mol of sulfuric acid dissolve in 50.0*mL of solution. What is the concentration with respect to "sulfuric acid"?

May 1, 2017

The first step (AS ALWAYS) is to write a stoichiometric equation......and I think your friend is right.

#### Explanation:

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

This is an acid base reaction, and as you know sulfuric acid is diprotic. And this is probably the source of the confusion: do I divide by 2; or do I multiply? We have all made these sorts of mistakes.

$\text{Moles of KOH} = 38.74 \times {10}^{-} 3 L \times 0.500 \cdot m o l \cdot {L}^{-} 1 = 19.37 \times {10}^{-} 3 \cdot m o l$.

Given the stoichiometric equation, there were $\frac{19.37 \times {10}^{-} 3 \cdot m o l}{2}$, i.e. half an equiv of sulfuric acid in the original solution.

And thus ${\text{Concentration}}_{{H}_{2} S {O}_{4}} = \frac{19.37 \times {10}^{-} 3 \cdot m o l}{2 \times 50.00 \times {10}^{-} 3 \cdot L}$

$= 0.194 \cdot m o l \cdot {L}^{-} 1$ with respect to sulfuric acid.

Note that normally we would titrate a volume of $50.00 \cdot m L$ with $38.74 \cdot m L$ titrant, not the reverse.