A30.00-ml sample of hydrochloric acid solution requires 28.23 ml of o.186 m sodium hydroxide for complete neutralization what is the concentration of the original hydrochloric acid solution? (30)

0.1750 M

Explanation:

The neutralization reaction is given by the chemical equation:

⇒

It means that for each NaOH molecule that reacts, need just one of HCl that react as well.

So, let's calculate the moles of NaOH that we needed for complete neutralization:

We know that the NaOH solution is 0.186M and we spent 28.23mL of this, so the total amount of moles can be calculated as:

That process we did was because the concentration is a rate of content (moles) per volume unit (L), and if we know the total volume needed, we can know the mole's quantity through the concentration.

Then, we know that 0.005251 mol NaOH were required, and the chemical equation says us that each NaOH molecule (or mol) reacts with just one of HCl, so, in the reaction, 0.005251 mol of HCl must have reacted. That 0.005251 moles were initially in a 30.00 mL volume, so, we can build the moles/volume rate, which is the concentration:

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%