I am doing a titration lab and need help with a problem. (i used .100M H2SO4)(I used 10.0 ml of Ba(OH)2 solution and added 50.0 ml of distilled water making 60.0 ml total.)
1. Use the titration results to calculate the moles of H2SO4 that were used to reach the equivalence point.
I found my equivalence point to be 8mL. The molar amount of H2SO4 therefore is 8e-4 moles.
2. Use your titration results to calculate the molar concentration (molarity) of the Ba(OH)2 solution using the molar amount of H2SO4.
Please help with question 2.
H2SO4 + Ba(OH)2 ==> BaSO4 + 2H2O
mols H2SO4 = M x L = 8E-4. right.
Now find mol Ba(OH)2 titrated.
8E-4 mols H2SO4 x (1 mol Ba(OH)2/1 mol H2SO4) = 8E-4 mols Ba(OH)2.
M Ba(OH)2 = mols Ba(OH)2/L Ba(OH)2
M Ba(OH)2 = 8E-4/0.010L = ?
Sure, I'd be happy to help with question 2!
To calculate the molar concentration of the Ba(OH)2 solution, you can use the balanced chemical equation between H2SO4 and Ba(OH)2:
2 H2SO4 + Ba(OH)2 -> BaSO4 + 2 H2O
From the balanced equation, we can see that the stoichiometric ratio between H2SO4 and Ba(OH)2 is 2:1.
Given that you used 8e-4 moles of H2SO4 at the equivalence point, this means that you would have used half the amount of Ba(OH)2, which is 4e-4 moles.
Now, we know the volume of the Ba(OH)2 solution used was 10.0 mL + 50.0 mL = 60.0 mL = 0.0600 L.
To calculate the molar concentration (molarity) of the Ba(OH)2 solution, we divide the moles of Ba(OH)2 by the volume in liters:
Molarity of Ba(OH)2 = moles of Ba(OH)2 / volume of solution (in L)
Molarity of Ba(OH)2 = 4e-4 moles / 0.0600 L = 6.67 M
So, the molar concentration of the Ba(OH)2 solution is approximately 6.67 M.
Remember, this is just a hypothetical scenario based on the information you provided. Make sure to double-check your calculations and use your actual measured data for accurate results.
To calculate the molar concentration (molarity) of the Ba(OH)2 solution, you first need to calculate the moles of Ba(OH)2 used in the reaction.
Since the balanced equation for the reaction is 1 mole of H2SO4 reacts with 2 moles of Ba(OH)2, we can use stoichiometry to find the moles of Ba(OH)2.
From the given information, we know that the moles of H2SO4 used is 8 x 10^-4 moles.
Using the stoichiometric ratio, we can calculate the moles of Ba(OH)2:
Moles of Ba(OH)2 = (moles of H2SO4) x (2 moles of Ba(OH)2 / 1 mole of H2SO4)
Moles of Ba(OH)2 = (8 x 10^-4 moles) x (2 moles of Ba(OH)2 / 1 mole of H2SO4)
Moles of Ba(OH)2 = 1.6 x 10^-3 moles
Next, we need to find the volume of the Ba(OH)2 solution used in the titration.
Given that 10 mL of Ba(OH)2 solution was used, we can calculate:
Volume of Ba(OH)2 solution (in liters) = 10 mL / 1000 mL/L
Volume of Ba(OH)2 solution = 0.01 L
Now, we can calculate the molar concentration (molarity) of the Ba(OH)2 solution by dividing the moles of Ba(OH)2 by the volume of the solution in liters:
Molarity of Ba(OH)2 solution = (moles of Ba(OH)2) / (volume of Ba(OH)2 solution in liters)
Molarity of Ba(OH)2 solution = (1.6 x 10^-3 moles) / (0.01 L)
Molarity of Ba(OH)2 solution = 0.16 M
Therefore, the molar concentration (molarity) of the Ba(OH)2 solution is 0.16 M.
To calculate the molar concentration (molarity) of the Ba(OH)2 solution using the molar amount of H2SO4, you need to use the balanced chemical equation and stoichiometry.
From the balanced equation of the reaction between H2SO4 and Ba(OH)2:
H2SO4 + 2Ba(OH)2 → BaSO4 + 2H2O
We can observe that the stoichiometric ratio between H2SO4 and Ba(OH)2 is 1:2. This means that for every 1 mole of H2SO4 used, 2 moles of Ba(OH)2 will react.
Given that the molar amount of H2SO4 used to reach the equivalence point is 8e-4 moles, we can now use stoichiometry to calculate the moles of Ba(OH)2 that reacted.
Since the stoichiometric ratio of H2SO4 to Ba(OH)2 is 1:2, we can multiply the moles of H2SO4 by 2 to get the moles of Ba(OH)2:
8e-4 moles H2SO4 * 2 moles Ba(OH)2/1 mole H2SO4 = 1.6e-3 moles Ba(OH)2
Next, we use the volume of the Ba(OH)2 solution to calculate the molar concentration (molarity). In the lab, you used 10.0 ml of Ba(OH)2 solution and diluted it to a total volume of 60.0 ml. We need to convert this to liters for calculation.
10.0 ml = 10.0 ml * (1 L/1000 ml) = 1.0 x 10^-2 L
Similarly, the total volume of the solution is 60.0 ml = 60.0 ml * (1 L/1000 ml) = 6.0 x 10^-2 L
Now, we can calculate the molar concentration (molarity) of the Ba(OH)2 solution:
Molarity (M) = moles/volume in liters
Molarity = 1.6e-3 moles Ba(OH)2 / 1.0 x 10^-2 L
Molarity = 1.6e-3 moles Ba(OH)2 / 1.0e-2 L
Molarity = 0.16 M Ba(OH)2
Therefore, the molar concentration (molarity) of the Ba(OH)2 solution is 0.16 M.