INFORMATION:
[HCl] = 0.1388 M
We placed 0.5 g Ca(OH)2 in 100 mL of the following solutions:
A 25 ml aliquot was used
Flask A: Distilled Water (7.4 mL of HCl needed to titrate)
Flask B: 0.05 M NaOH (7.5 mL of HCl needed to titrate)
Flask C: 0.025 M NaOH (10.8 mL of HCl needed to titrate)
Flask D: 0.0125 M NaOH (8.9 mL of HCl needed to titrate)
QUESTION is to calculate the following for each flask:
[OH-]eq from titration data
[OH-]0 from original solution
[OH-] from Ca(OH2)
[Ca2+]eq from Ca(OH)2
Ksp = [Ca2+]eq * [OH-]eq
I got moles HCl by volume (L) * concentration (0.1388M)
Is the [OH-]eq the same as the moles of HCl?
I must confess I don't understand this at all.
0.5 g of Ca(OH)2 was placed in a flask. 100mL of 0.05M NaOH was poured into the flask. 25mL aliquot was filtrated and used for titration.
Suppose to calculate the OH^- equilibrium concentration from the titration data, as well as the OH^- from the original solution. Then calculate the OH^- from Ca(OH)2 and then calculate Ca^2+ from Ca(OH)2.
If I can get an example solution for the above I can calculate the rest/
Thank You!
No, the [OH-]eq is not the same as the moles of HCl. The [OH-]eq refers to the equilibrium concentration of hydroxide ions in the solution, which is determined by the amount of HCl required to neutralize the reactants in the flask. On the other hand, the moles of HCl you calculated can be used to determine the moles of hydroxide ions present in the original solution.
To calculate [OH-]eq from the titration data, you need to use the volume of HCl needed to titrate each flask. Given that 1 mole of HCl reacts with 1 mole of hydroxide ions (OH-), you can use the balanced chemical equation to determine the moles of OH- present in each flask.
For example, in Flask A, 7.4 mL of HCl was needed to titrate the solution. Using the known concentration of HCl (0.1388 M), you can convert the volume to moles by multiplying the volume (in liters) by the concentration. This would give you the moles of HCl used. Since 1 mole of HCl reacts with 1 mole of hydroxide ions, the moles of HCl used would also represent the moles of OH- present in the solution. Therefore, [OH-]eq for Flask A would be equal to 0.1388 moles/L.
To calculate [OH-]0 from the original solution, you need to consider the dilution that occurred when adding 0.5 g of Ca(OH)2 to 100 mL of the solution. To do this, you can use the volume of the aliquot taken (25 mL) and the total volume of the original solution (100 mL) to calculate the dilution factor. Then, you multiply the [OH-]eq by the dilution factor to obtain [OH-]0.
The [OH-] from Ca(OH)2 refers to the concentration of hydroxide ions that ionize from the dissociation of calcium hydroxide in water. To calculate this, you can use the molar mass of Ca(OH)2 to determine the number of moles of Ca(OH)2 dissolved in the 25 mL aliquot. Since calcium hydroxide has a 1:2 stoichiometric ratio with hydroxide ions, you can multiply the moles of Ca(OH)2 by 2 to obtain the moles of OH-. Then, you can divide this by the total volume of the aliquot (25 mL) to get [OH-] from Ca(OH)2.
To calculate [Ca2+]eq from Ca(OH)2, you need to use the balanced chemical equation for the dissociation of calcium hydroxide. From the equation, you can determine that for every 1 mole of calcium hydroxide that dissolves, 1 mole of calcium ions (Ca2+) are formed. Therefore, the concentration of Ca2+ would be equal to the concentration of OH- from Ca(OH)2.
Lastly, to calculate Ksp, you multiply [Ca2+]eq by [OH-]eq, as Ksp is the equilibrium constant expression for the dissolution reaction of calcium hydroxide.
Repeat these calculations for each flask using the corresponding titration data and concentrations, and you will obtain the results for each of the mentioned parameters.