# What is the molarity of a sample of sulfuric acid if 28.00 mL of 0.18 M NaOH was required to titrate 10 mL of the sulfuric acid?

##### 1 Answer
Jun 9, 2017

Approx. $0.25 \cdot m o l \cdot {L}^{-} 1. \ldots \ldots \ldots . .$ with respect to ${H}_{2} S {O}_{4}$

#### Explanation:

We need (i) a stoichiometric equation:

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

And (ii) equivalent quantities of sodium hydroxide.........

$\text{Moles of NaOH} = 28.00 \times {10}^{-} 3 L \times 0.18 \cdot m o l \cdot {L}^{-} 1 = 5.040 \times {10}^{-} 3 \cdot m o l$

And thus [H_2SO_4]=(1/2xx5.040xx10^-3*mol)/(10xx10^-3L)=??