A 20.0 Ml sample of sulfuric acid is titrated with 25.0 ML of 1.0 M NaOH solution. What is the molarity of the sulfuric acid? H2SO4+ 2NaOH yields Na2SO4 + H2O
molesacid=2molesbase
20mL*Molarity=2*1M*25mL
molaritacid=2*25/20=2.5M
To determine the molarity of the sulfuric acid, we can use the balanced chemical equation and the concept of stoichiometry.
Step 1: Write the balanced chemical equation:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
Step 2: Determine the stoichiometry of the reaction:
From the equation, we can see that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide. This means that the mole ratio between sulfuric acid and sodium hydroxide is 1:2.
Step 3: Calculate the number of moles of NaOH used:
To calculate the number of moles of sodium hydroxide used, we can use the molarity and volume of the NaOH solution:
Moles of NaOH = Molarity × Volume
= 1.0 M × 25.0 mL
= 25.0 mmol (millimoles)
Step 4: Use the stoichiometry to determine the number of moles of sulfuric acid:
Since the mole ratio of sulfuric acid to sodium hydroxide is 1:2, we can conclude that the number of moles of sulfuric acid is half the number of moles of sodium hydroxide used.
Moles of H2SO4 = 1/2 × Moles of NaOH
= 1/2 × 25.0 mmol
= 12.5 mmol
Step 5: Calculate the molarity of the sulfuric acid:
Now, we need to convert the moles of sulfuric acid to the corresponding volume in liters:
Volume of H2SO4 (L) = Moles of H2SO4 / Molarity of H2SO4
Since the volume of sulfuric acid is given as 20.0 mL, we need to convert it to liters before substituting the values:
Volume of H2SO4 (L) = 20.0 mL × (1 L/1000 mL)
= 0.020 L
Now, we substitute the values into the equation:
0.020 L = 12.5 mmol / Molarity of H2SO4
Rearranging the equation, we can solve for the molarity of sulfuric acid:
Molarity of H2SO4 = Moles of H2SO4 / Volume of H2SO4
= 12.5 mmol / 0.020 L
= 625 M
Therefore, the molarity of the sulfuric acid is 625 M.