Question

In a certain experiment, 28.0 mL of 0.250 M HNO3and 53.0 mL of 0.320 M KOH are mixed. Calculate the number of moles of water formed in the resulting reaction. What is the concentration of H+ or OH− ions in excess after the reaction goes to completion?

Answer 1

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