We are back to sodium experiments, this time in much more controlled conditions. What is percent concentration of NaOH in solution prepared by adding 4.00 g of metallic sodium to 50.0 mL of water?

We have to calculate two things - mass of sodium hydroxide and mass of the solution. 50 mL of water weights 50 g (not exactly and it will depend on the temperature, but for our purposes that's enough accurate assumption) which is probably excess high enough to assume that Na is a limiting reagent here.

2Na | + | 2H_{2}O | → | 2NaOH | + | H_{2} |

45.98 g | : | 79.99 g | ||||

4.00 g | : | x g |

So proportion is

45.98 g : 79.99 g = 4.00 g : x g

and x = 75.99/45.98×4.00 = 6.96 g

Note that this mass is small enough to confirm our assumption about water excess. Product weights about three grams more than sodium used, this mass increase means we used only about 3 g of water for the reaction.

Now we need mass of the solution - and that's the place were it is easy to overlook one important fact - hydrogen evolved and left, thus final mass is 50.0 g (water) + 4.00 g (Na) - unknown yet mass of hydrogen. But that's not difficult to calculate:

2Na | + | 2H_{2}O | → | 2NaOH | + | H_{2} |

45.98 g | : | 2.02 g | ||||

4.00 g | : | x g |

Proportion is

45.98 g : 2.02 g = 4.00 g : x g

and x = 2.02/45.98×4.00 = 0.175 g

So final mass of the solution is 50.0 + 4.00 - 0.175 g = 53.83 g

All we have to do now is to insert known values into percent concentration definition, to obtain

C_{%w/w} = 6.96/53.83 × 100% = **12.9%**