# Question 9c03b

Feb 19, 2015

The molarity of the sulfuric acid solution will be $\text{17.1 M}$.

You are dealing with a neutralization reaction - in this case, a strong acid and a strong base react to produce salt and water. The balanced chemical equation looks like this

$2 N a O H + {H}_{2} S {O}_{4} \to N {a}_{2} S {O}_{4} + 2 {H}_{2} O$

SIDE NOTE - the reaction takes place in aqueous solution; taht being said, I won't go into the net ionic equation

Now look at the mole ratio you have between $N a O H$ and ${H}_{2} S {O}_{4}$ - 2 moles of the former react with 1 mole of the latter. A neutralization reaction implies that both the acid and the base are consumed, i.e. the two cancel each other out.

This means that the number of ${H}_{2} S {O}_{4}$ moles depends on the number of moles of $N a O H$, which you can determine by using its molar mass, and on the mole ratio between the two compounds

$\text{112.0 g" * ("1 mole")/("40.0 g") = "2.80 moles NaOH}$

Apply the aforementioned mole ratio to see exactly how many moles of ${H}_{2} S {O}_{4}$ were neutralized by this much $N a O H$

$\text{2.80 moles NaOH" * ("1 mole"H_2SO_4)/("2 moles NaOH") = "1.40 moles}$ ${H}_{2} S {O}_{4}$

The last thing to do is use the formula for molarity

C = n/V = ("1.40 moles")/(82.0*10^(-3)"L") = "17.1 M"#

Don't forget to transform mL into L, since molarity uses liters, not mililiters!

Feb 19, 2015

First convert the weight of $N a O H$ to moles.

Relative atomic masses are $N a = 23 , O = 16 , H = 1$
For a total molecular mass $N a O H = 40$

So $112.0 g$ converts to $\frac{112.0}{40} = 2.8 m o l$

Since sulfuric acid ${H}_{2} S {O}_{4}$ has two ${H}^{+}$-ions it neutralises in a ratio of 2:1, so there must have been $2.8 / 2 = 1.4 m o l$ of sulfuric acid.

The molarity of the solution= $\frac{m o l}{L} .$

Molarity = $\frac{1.4 \text{mol}}{82.0 m L} = \frac{1.4 m o l}{0.082 L} \approx 17.1 m o l / L$

Which is a very strong solution -- you may (almost) call this concentrated sulfuric acid. Don't spill it on your skin!

Extra :
Let's call this a thought experient. You would NOT want to do this in practice, not even in a fume cupboard, as the result will get explosive due to the extreme heat developed in this reaction!