Please be specific!
Briefly explain whether each if the following procedural errors would result in a high, low, or an unchanged calculated mass of Ca2+ in the analyzed tablet.
a)a student had difficulty decanting the supernatant solution from the test tubes in step 9, causing him to transfer some of the solid along with the solution.
b)when rinsing the binder material from the test tubes, student observed small pieces of undissolved tablet in the solid at the bottom of the tubes.
c)a student found that the litmus paper did not turn blue when tested with a drop of the solution, then failed to add additional Na_2_CO_3_ solution and test the solution again beforeproceeding to the next step.
d)a student discarded the torn-off filter paper corner, instead of placing it in the cone
e)after labeling the filter paper, a student left wet fingerprints on the paper when transferring it to the balance to determine its mass.
f)a student failed to determine the mass of the filter paper prior to filtering the CaCO_3_. Realizing the importance of this mass, he selected a new piece of filter paper and used its mas in his calculations.
2)Briefly expalin why you can use this experiment's procedure to determine the number of milligrams of Ca2+ ion in a food sample in which the Ca2+ ion is present in a form ofther than CaCO_3_
2 answers
Write an equation for each step. Substitute (mentally) high or low for the part of the equation that is changing, determine the effect at that stage, then go to the next step. Here is an example.
Doing an experiment for % Ca in lime (Ca(OH)2 in a sample by titrating with 0.1M HCl
2HCl + Ca(OH)2 ==> CaCl2 + 2H2O
1. mols HCl = mL x M
2. mols Ca(OH)2 = mols HCl/2
3. g Ca(OH)2 = mols Ca(OH)2 x molar mass Ca(OH)2.
4. %Ca(OH)2 = (mass Ca(OH)2/mass sample)*100 = ?
question: How will percent Ca(OH)2 be affected if we over titrate with HCl; i.e., add too much HCl.
Too much HCl in step 1 (mL too high) means mols too high. Go to step 2.
Too many mols HCl in step 1 means too many mols Ca(OH)2 in step 2. Go to 3.
Too many mols Ca(OH)2 means grams Ca(OH)2 will be too high and that substituted into 4 for mass Ca(OH)2 gives high results for percent Ca(OH)2.