25.00mL of an unknown sulfuric acid solution is titrated to the second equivalence point with 21.02mL of 0.420M potassium hydroxide solution. What is the concentration of the sulfuric acid solution?
H2SO4 + 2KOH ==> K2SO4 + 2H2O
mols KOH = M x L = ?
mols H2SO4 = 1/2 mol KOH (see the coefficients)
M H2SO4 = mols H2SO4/L H2SO4
To find the concentration of the sulfuric acid solution, we can use the concept of stoichiometry and the equation of the reaction between sulfuric acid and potassium hydroxide. The balanced equation is:
H2SO4 + 2KOH → K2SO4 + 2H2O
From the equation, we can see that one mole of sulfuric acid reacts with two moles of potassium hydroxide to produce one mole of potassium sulfate and two moles of water.
First, let's calculate the number of moles of potassium hydroxide that reacted:
moles of KOH = volume of KOH solution (in L) × concentration of KOH (in mol/L)
= 21.02 mL × (1 L/1000 mL) × 0.420 mol/L
= 0.0088 mol
Since the stoichiometric ratio between sulfuric acid and potassium hydroxide is 1:2, the number of moles of sulfuric acid is equal to half the number of moles of potassium hydroxide:
moles of H2SO4 = 0.0088 mol ÷ 2
= 0.0044 mol
Next, we need to calculate the concentration of the sulfuric acid solution:
concentration of H2SO4 = moles of H2SO4 ÷ volume of sulfuric acid solution (in L)
= 0.0044 mol ÷ 25.00 mL × (1 L/1000 mL)
= 0.176 mol/L
Therefore, the concentration of the sulfuric acid solution is 0.176 mol/L.