What volume of 0.29M H2SO4 is required to neutralize 50mL of 0.29M Ba(OH)2 solution?
The balanced chemical equation for the reaction between sulfuric acid (H2SO4) and barium hydroxide (Ba(OH)2) is:
H2SO4 + Ba(OH)2 -> BaSO4 + 2H2O
From the equation, we can see that the mole ratio between H2SO4 and Ba(OH)2 is 1:1.
To calculate the volume of 0.29M H2SO4 required to neutralize 50mL of 0.29M Ba(OH)2 solution, we can use the formula:
M1V1 = M2V2
Where:
- M1 is the molarity of the H2SO4 solution
- V1 is the volume of the H2SO4 solution
- M2 is the molarity of the Ba(OH)2 solution
- V2 is the volume of the Ba(OH)2 solution
Since the mole ratio is 1:1, the volume of H2SO4 required is equal to the volume of Ba(OH)2 solution:
V1 = V2
We can rearrange the formula to solve for V1:
V1 = M2V2 / M1
V1 = (0.29 mol/L)(0.050 L) / 0.29 mol/L
V1 = 0.050 L
Therefore, 50mL of 0.29M H2SO4 is required to neutralize 50mL of 0.29M Ba(OH)2 solution.