I carefully measured 3.0 mL of 0.045MCa (NO_3)_2 and placed it in a test tube wi

Question

I carefully measured 3.0 mL of 0.045MCa (NO_3)_2 and placed it in a test tube with 5.0 mL of 0.045 M NaF. The formula for the resulting precipitate is: _____ I added warm water to the mixture above until the precipitate disappeared. The amount of water required to dissolve all precipitate was 13.0 mL. What is the experimentally determined value for Ksp the precipitate based on these results? The literature value of Ksp for this particular ionic compound is 1.8 times 10^-7 What factors in my experiment could explain the difference between the literature value and the experimentally determined value for Ksp above

Explanation / Answer

4) Mixing Ca(NO3)2 and NaF

Ca(NO3)2(aq) + 2NaF(aq) ---> CaF2 (ppt) + 2NaNO3(aq)

Formula of precipitate = CaF2

moles Ca2+ = 0.045 M x 3 ml = 0.135 mmol

moles F- = 0.045 ml x 5 ml = 0.225 mmol

moles ppt = 0.1125 mmol

[CaF2] formed = 0.1125 mmol/8 ml = 0.0141 M

Ksp = [Ca2+][F-]^2

       = (0.0141)(2 x 0.0141)^2

       = 1.12 x 10^-5 M

The difference between literature Ksp and calcuated Ksp could be due to lower concentration of Ca2+ and F- taken initially, dilution increases solubility of solid formed.

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