Calculations and Results: lab K sp of ca(oh)2 Method 1: Using the titration curv

Question

Calculations and Results: lab Ksp of ca(oh)2

Method 1: Using the titration curve.

Determine the moles of acid used to reach equivalence point

method 2:

Determine the number of moles of hydroxide neutralized by the acid.

My question is:

   Parts i-iv contain errors in that could occur in today’s lab.  Explain whether each would cause your calculated Kspvalue to be higher or lower or remain unchanged to the tabulated value. Assume you are using method 1.

i. The calcium hydroxide solution was prepared at the last minute! Adequate time was not given for equilibrium to be reached between the solid and dissolved calcium hydroxide.

ii. The burette was not rinsed with 10 mL of HCl and HCl was not run through the stopcock before it was filled with acid – thanks lab partner! Therefore, some DI water was on the sides of the burette and within the stopcock.

iii. The pH meter was not calibrated before the experiment and all the pH readings were 0.2 pH units higher than reality.

iv. The HCl was not diluted correctly. Although labeled 0.020 M, it was really 0.030 M.  

Explanation / Answer

Ksp = [A +] * [B-]

i) The solution of calcium hydroxide has not reached equilibrium, therefore the concentration of ions of that salt have not been completely released (they are less) since Ksp is directly proportional to the concentration of ions, if there is not enough the Ksp It will be less.

ii) Water is the solvent of both ions, so if there is excess water, the concentration of both ions decreases, but it does so proportionally, so it does not affect the Ksp.

iii) Does not affect Ksp, but other values ??calculated in the laboratory.

iv) If more moles of HCl are added than stipulated, it will also increase the dissolved ions in the solution, increasing the Ksp.

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