If 38.33 mL of a standard 0.1667 M Ba(OH)2 solution is required to neutralize exactly 52 mL of H2SO4, what is the molarity of the acid solution?
Ba(OH)2 + H2SO4 ==? BaSO4 + 2H2O
moles Ba(OH)2 = M x L = ?
Convert moles Ba(OH)2 to mol H2SO4.
?mol Ba(OH)2 x [1 mol H2SO4/1 mol Ba(OH)2]= ?mol Ba(OH)2 x 1/1 which means moles Ba(OH)2 = mol H2SO4.
Then M H2SO4 = moles H2SO4/L H2SO4.
To find the molarity of the acid solution, we can use the equation:
Molarity of acid solution (H₂SO₄) * Volume of acid solution (H₂SO₄) = Molarity of base solution (Ba(OH)₂) * Volume of base solution (Ba(OH)₂)
Given:
Volume of base solution (Ba(OH)₂) = 38.33 mL
Molarity of base solution (Ba(OH)₂) = 0.1667 M
Volume of acid solution (H₂SO₄) = 52 mL
Let's substitute these values into the equation:
Molarity of acid solution (H₂SO₄) * 52 mL = 0.1667 M * 38.33 mL
Now, solve for the molarity of the acid solution:
Molarity of acid solution (H₂SO₄) = (0.1667 M * 38.33 mL) / 52 mL
Calculating this expression, we get:
Molarity of acid solution (H₂SO₄) = 0.122246 M
Therefore, the molarity of the acid solution is approximately 0.122 M.