# What volume of 18*mol*L^-1 H_2SO_4(aq) is needed to make a 250*mL volume of solution that is 0.50*mol*L^-1 with respect to H_2SO_4?

Aug 24, 2017

This is where you use a dilution factor/equation to prepare 250 mL of 0.50 M ${H}_{2} S {O}_{4} \left(a q\right)$. This is the ${M}_{1} {V}_{1} = {M}_{2} {V}_{2}$.

In this case:
${M}_{1} = 18 M$
${M}_{2} = 0.50 M$
${V}_{2} = 250 m L$

Note that when you use the dilution equation, there is no need to switch from mL to L. Let's solve for ${V}_{1}$.

${V}_{1} = \frac{{M}_{2} {V}_{2}}{M} _ 1 = \frac{\left(0.50 M\right) \left(250 m L\right)}{18 M} = 6.944 m L$

Due to sig figs, your answer should be 6.9 mL. You need to add 6.9 mL of 18M sulfuric acid and dilute with water until you reach 250 mL.

Aug 24, 2017

#### Explanation:

You add acid to water because if you spit in conc. acid it spits back (I kid you not!), and when you do the dilution you wear safety spex and a lab coat to protect your clothing. I write this first because this is an ABSOLUTE SAFETY REQUIREMENT not only for yourself but for the people around you. You want to finish the experiment with your mince pies intact.

We wants a $250 \cdot m L$ volume of $0.50 \cdot m o l \cdot {L}^{-} 1$ ${H}_{2} S {O}_{4}$.

And this corresponds to a molar quantity of $250 \times {10}^{-} 3 \cdot L \times 0.50 \cdot m o l \cdot {L}^{-} 1 = 0.125 \cdot m o l$ WITH RESPECT TO SULFURIC ACID.

And we get this from the conc. sulfuric acid.....

$\frac{0.125 \cdot m o l}{18.0 \cdot m o l \cdot {L}^{-} 1} \times {10}^{3} \cdot m L \cdot {L}^{-} 1 = 6.94 \cdot m L$.

And so we add slightly under $7 \cdot m L$ of conc. acid to approx. $240 \cdot m L$ of water. Again I stress that this order of addition is absolutely important.