# What is the molarity of an H2SO4 solution if 18.5 mL of 0.18 M NaOH are needed to neutralize 25.0 mL of the sample? H2SO4 + 2NaOH => Na2SO4 + 2H2O

Apr 16, 2015

The molarity is $0.066 \text{mol/l}$

$c = \frac{n}{v}$

$n = c v$

So we can work out the no. moles $N a O H$:

$n N a O H = \frac{18.5 \times 0.18}{1000} = 3.33 \times {10}^{- 3}$

From the equation we can see that the no. moles ${H}_{2} S {O}_{4}$ must be half that amount so:

$n {H}_{2} S {O}_{4} = \frac{3.33}{2} \times {10}^{- 3} = 1.66 \times {10}^{- 3}$

$c = \frac{n}{v}$ so:

$\left[{H}_{2} S {O}_{4}\right] = \frac{1.66 \times {10}^{- 3}}{25.0 \times {10}^{- 3}} = 0.066 \text{mol/l}$