# A 25*mL volume of sodium hydroxide solution of 0.150*mol*L^-1 concentration is stoichiometrically equivalent to a 15.0*mL of sulfuric acid. What is [H_2SO_4]?

Feb 24, 2017

To find $\left[{H}_{2} S {O}_{4}\right]$, we need (i), a stoichiometrically balanced equation:

#### Explanation:

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

And (ii) equivalent quantities of sodium hydroxide,

$\text{Moles of NaOH} = 25.0 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.150 \cdot m o l \cdot {L}^{-} 1 = 3.75 \times {10}^{-} 3 \cdot m o l$

Given the equation, we KNOW that there was a half an equiv of sulfuric acid present in the original volume of ${H}_{2} S {O}_{4} \left(a q\right)$.

And thus $\left[{H}_{2} S {O}_{4}\right] = \frac{\frac{1}{2} \times 3.75 \times {10}^{-} 3 \cdot m o l}{15.0 \times {10}^{-} 3 L} \cong 0.13 \cdot m o l \cdot {L}^{-} 1$

All I have done here is to use the relationship:

$\text{Concentration}$ $\equiv$ $\text{Moles of solute"/"Volume of solution.}$

And thus,

$\text{Concentration"xx"Volume"="Moles of solute}$, etc. etc...........