Answer:
The number of moles of water formed in the resulting reaction is 6.03
[H+]: 37,2 M
[OH-]: 37,2 M
Explanation:
HNO3 + KOH ----> KNO3 + H2O
First, we must discover the limiting reagent and we need to find out the moles, we use for this.
Moles that are used = Molarity / volume
HNO3 : 0,250 mol/L / 0,028L = 8,93 moles
KOH : 0,320 mol/L / 0,053L = 6,03 moles
The ratio of the reagents by stoichiometry is 1 to 1, so the limiting reagent is KOH, if I need 1 mole of nitric per mole of KOH, for every 8.93 moles I will need the same. However I have only 6.03 moles of KOH
The ratio of the reagents/products by stoichiometry is 1 to 1 so if I need 1 mol of KOH to make 1 mol of Water, 6,03 moles of KOH are used to make 6,03 moles of H2O.
The equilibrium of water is this:
2H2O ⇄ H3O+ + OH-
2 moles of water are broken down into 1 mole of hydronium (H3O +) and 1 mole of hydroxyl (OH-)
6,03 moles of water are broken down into the half of those moles, so we have 3,015 moles of H3O+ and 3,015 moles of OH- but these moles are in 81,0 mL (the volume of the two solutions, 28 mL + 53 mL)
We must find out the moles in 1000 mL (1 L) so let's apply the rule of three.
81 mL ____ 3,015 moles
1000 mL ___ ( 1000 . 3,015) /81 = 37,2 M
