Chemistry Challenge: Calculating the Moles of Sodium Hydroxide

How can we calculate the moles of sodium hydroxide in a mixture of sodium hydroxide and strontium hydroxide?

We have a mixture of sodium hydroxide and strontium hydroxide with a total of 0.19 moles of the two compounds. It takes 100.0 mL of 2.969 M HCl to neutralize all of the base. How many moles of sodium hydroxide were in the original mixture?

Answer:

0.0831 moles of NaOH were in the original mixture.

The calculation involves setting up equations to determine the moles of sodium hydroxide and strontium hydroxide in the original mixture. By understanding the stoichiometry of the neutralization reaction and using the volume and concentration of HCl, we can accurately find the moles of each compound.

First, we need to write two equations to represent the total moles of each compound and the moles of HCl required to neutralize them:

(1) 0.19mol = X + Y

Where X = Moles NaOH; Y = Moles Sr(OH)2

The moles of HCl required can be calculated using the volume and concentration:

0.100L * (2.969mol / L) = 0.2969 moles HCl

We can then set up a second equation using the moles of HCl and the moles of sodium hydroxide and strontium hydroxide:

(2) 0.19mol - Y = X

0.2969 mol = (0.19mol - Y) + 2Y

Solving these equations, we find that the moles of sodium hydroxide in the original mixture is 0.0831 moles.

This calculation showcases the importance of stoichiometry and careful consideration of the chemical reactions involved in determining the composition of a mixture.