If 42.68 ml of 0.43 M KOH is used to neutralize 18.40 ml of H, SO4, calculate the concentration of the sulfuric acid solution.
Write the net ionic equation of this reaction

The concentration of the sulfuric acid : 0.499 M

The net ionic equation

2 OH ⁻(aq]+ 2 H ⁺ (aq] → 2 H ₂O (l]

Given

42.68 ml of 0.43 M KOH

18.40 ml of H2SO4

Required

the concentration of the sulfuric acid

the net ionic equation

Solution

Acid-base titration formula  

Ma. Va. na = Mb. Vb. nb  

Ma, Mb = acid base concentration  

Va, Vb = acid base volume  

na, nb = acid base valence  

Input the value :

a=KOH, b= H2SO4

0.43 x 42.68 x 1 = Mb x 18.40 x 2

Mb = 0.499 M

The net ionic equation

Reaction

2KOH + H2SO4 → K2SO4 + 2H2O

2 K ⁺ (aq]+ 2 OH ⁻ (aq] + 2 H ⁺ (aq] + SO ₄²⁻(aq] → 2 K ⁺ (aq] + SO ₄²⁻(aq] + 2 H ₂ O (l]

canceled the spectator ions :

2 OH ⁻(aq]+ 2 H ⁺ (aq] → 2 H ₂O (l]

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 :