A volume of 80.0 mL of aqueous potassium hydroxide (KOH) is titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 12.7 mL of 1.50 M H2SO4 was needed

0.478 M

Explanation:

Let's consider the neutralization reaction between KOH and H₂SO₄.

2 KOH + H₂SO₄ → K₂SO₄ + 2 H₂O

12.7 mL of 1.50 M H₂SO₄ react. The reacting moles of H₂SO₄ are:

0.0127 L × 1.50 mol/L = 0.0191 mol

The molar ratio of KOH to H₂SO₄ is 2:1. The reacting moles of KOH are 2 × 0.0191 mol = 0.0382 mol

0.0382 moles of KOH are in 80.0 mL. The molarity of KOH is:

M = 0.0382 mol/0.0800 L = 0.478 M

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :