Chemistry - Solubility
posted by PleaseHelpThankYou! .
[HCl] = 0.1388 M
We placed 0.5 g Ca(OH)2 in 100 mL of the following solutions:
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?
Chemistry - Solubility -
Also a 25 ml aliquot was used.