A 31.5 mL aliquot of aqueous sulfuric acid of unknown concentration was titrated with
0.0134 M aqueous sodium hydroxide. It took 23.9 mL of the base to reach the endpoint of the
titration. The concentration (M) of the acid was

Question

A 31.5 mL aliquot of aqueous sulfuric acid of unknown concentration was titrated with
0.0134 M aqueous sodium hydroxide. It took 23.9 mL of the base to reach the endpoint of the
titration. The concentration (M) of the acid was

matt101 · Answer

First let's look at the reaction:

$$H_2SO_4+2NaOH \rightarrow 2H_2O+Na_2SO_4$$
Since H2SO4 produces two equivalents of acid, it takes twice as many moles of the base to fully neutralize it. This means that if we know how much base was used, we can figure out how much acid there was.

For the base, NaOH, we have a concentration (C) and we have a volume (V). This let's us find moles (n) by:

$$C={n \over V}$$$$0.0134={n \over 0.0239}$$$$n=3.2 \times 10^{-4}$$
So we have 3.2 x 10^(-4) mol of NaOH that reacted. Based on our equation above, we know that there are half as many moles, or 1.6 x 10^(-4) mol, of H2SO4 that reacted. We're given the volume of the H2SO4 solution, so we can now find its concentration:

$$C={n \over V}$$$$C={1.6 \times 10^{-4} \over 0.0315}$$$$C=5.1 \times 10^{-3}$$
The concentration of the unknown H2SO4 solution was therefore 5.1 x 10^(-3) M.

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