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Chemistry - Solubility

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[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?

  • Chemistry - Solubility -

    I must confess I don't understand this at all.

  • Chemistry - Solubility -

    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!

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