The Determination of the Solubility of Calcium Hydroxide
I have to plan an experiment to find the solubility of calcium
hydroxide, Ca(OH)2, in water. I have to make up a solution of calcium
hydroxide and carry out a titration using hydrochloric acid solution
of the chosen concentration.
The equipment need is as below:
· Solid calcium hydroxide
· Methyl orange indicator
· Volumetric flask (250cm3)
· Clamp and boss
· Clamp stand
· Burette (50cm3)
· Conical flask
· Pipette (25cm3)
· Pipette filler
· Distilled water
· White spotting tile
· Hydrochloric acid of chosen concentration
· Beaker x2
· Rubber bung
· Funnel x2
· Electronic scale
‘The maximum mass of calcium hydroxide needed to produce 1dm3 of
saturated solution at room temperature is 1.5g.’
I only want 250cm3 as I am using a 250cm3 volumetric flask. Therefore:
1dm3 / 4 = 250cm3
1.5g / 4 = 0.375g
The number of moles in volumetric flask:
0.375 / 74 = 0.005 moles
I need an excess of 0.5g to make sure that all the calcium hydroxide
has been fully dissolved:
0.375g + 0.5g = 0.875g
I have to now work out the concentration of hydrochloric acid I will
be using. The molar mass of calcium hydroxide is:
C = 40 O = 16 (x2) H = 1 (x2)
R.A.M = 74
The concentration of calcium hydroxide at the beginning will be:
1.5 / 74 = 0.02
So, the concentration is 0.02 mol/dm3
In the experiment I will be using 25cm3 of the solution from the
volumetric flask, so the mass of the calcium hydroxide in one
titration will be:
0.375 / 10 = 0.0375g
Therefore the number of moles of calcium hydroxide:
0.0375 / 74 = 0.0005 moles
Ca(OH)2(aq) + 2HCl à CaCl2(aq) + 2H20(l)
The purpose for this lab was to use aluminum from a soda can to form a chemical compound known as hydrated potassium aluminum sulfate. In the lab aluminum waste were dissolved in KOH or potassium sulfide to form a complex alum. The solution was then filtered through gravity filtration to remove any solid material. 25 mLs of sulfuric acid was then added while gently boiling the solution resulting in crystals forming after cooling in an ice bath. The product was then collected and filter through vacuum filtration. Lastly, crystals were collected and weighed on a scale.
After working through many calculations I came out with an average constant of 280, an accurate measurement. Although my readings caused me to have an accurate final answer, they were not precise. My values for the equilibrium constant varied greatly in some of ten trials, ranging from a low of 260 to a high of 320. Other contributions to the value of the constant would be the accuracy of the measuring devices, the purity of the solution and the accuracy of the best-fit line drawn on the graph. Since one of these solutions is clear and the other is colored their Concentrations can easily be found. The solutions can be simply put into a spectrometer and the absorbance will reveal how much of the colored solution resides in the solution. Your results in part one of the experiment can be used to create a graph with which you can make a best fit line and find values for the absorbencies in part two. This information can then be used to calculate the equilibrium constant in all or ten trials and an average can be taken. It allows the student to view first hand exactly what happens at equilibrium and then put this knowledge to
Well, this looks like its using some calculations so what I would do is take my 0.045 M and equal it to the 0.25 mL of NH3 and multiply that by 45.0 mL and multiply it by 10 with an exponent of negative 3. Once all of that is multiplied together we should get an answer of 0.01135 moles of our HCI. Now we can find our “Concentration” Which means we would divide our moles (0.01125) to our vol in liters which is 0.025, once we do that, we get an answer of 0.045M of our NH3. Well, since we are on the topic of pH we know that we can use the formula: pH = -log (H3O+). Then what we would do is plug everything into the formula: pH equals -log (2.4 multiplied by 10 (with an exponent of -5). Once we find the answer to this and we add up all of our calculations, we can come to a conclusion that the answer is: 4.6197 as our pH.
There are a few changes that should be made to the procedure so that the experiment could generate better results. More time should be allowed to dissolve the tablets as if they are not dissolved colour changes are harder to identify during the titration and also the results are less accurate. The acid used should also be more concentrated as a 0.5mol or 0.3mol would mean a smaller amount of acid would be required.
Volume's Effect on a Copper Sulphate Solution We are trying to find out if the current though a copper sulphate solutions volume is increased. To find this information out I shall perform an experiment using the following equipment; · 1 power pack · 1 beaker · 2 carbon rods for anode and cathode · 1 ammeter · 1 measuring cylinder · 2 crocodile clip wirers I shall also be using 60cm3 volume of copper sulphate in my preliminary results to decide upon the concentration of copper sulphate and the voltage I shall use. The following diagrams show the step by step process in which I will do my experiment; [IMAGE] [IMAGE] [IMAGE] [IMAGE] I will take 10 readings from 10cm3 to 100cm3. I will repeat my experiment to give my experiment a fair average. I will keep the power pack the beaker the carbon rods the crocodile clips the ammeter the concentration of copper sulphate and the measuring cylinder the same each time I do the experiment this experiment.
3. Add on of the following volumes of distilled water to the test tube, as assigned by your teacher: 10.0mL, 15.0mL, 20.0mL, 25.0mL, 30.0mL. (If you use a graduated cylinder, remember to read the volume from the bottom of the water meniscus. You can make more a more accurate volume measurement using either a pipette or a burette.)
Solubility is defined as the maximum amount of a substance that will dissolve in a given amount of another substance at constant temperature and pressure. Solubility is typically expressed in terms of maximum volume or mass of the solute that dissolve in a given volume or mass of a solvent. Traditionally the equilibrium solubility at a given pH and temperature is determined by the shake flask method. According to this method the compound is added in surplus to a certain medium and shaken at a predetermined time. The saturation is confirmed by observation of the presence of un-dissolved material. Saturation can also be reached if the solvent and excess solute is heated and then allowed to cool to the given temperature. After filtration of the
will result in an increase in the speed of the rate of reaction it has
Investigating the Rate of Reaction Between Marble Chips and the Varying Concentrations of Hydrochloric Acid
Investigate how the concentration of hydrochloric acid effects the rate at which it reacts with calcium carbonate
Investigating the Effect of Concentration of Dilute Hydrochloric Acid with Magnesium Metal Aim: To investigate the effect of concentration of dilute hydrochloric acid when it reacts with magnesium metal. Scientific Knowledge: Concentrated acid contains many acid particles. In order for a reaction to take place acid particles must collide with magnesium atoms breaking the chemical bonds, there must also be enough energy within the reaction for them to collide; otherwise they would simply bounce off each other. A reaction that does have enough energy to create a reaction is referred to as an effective collision. Within a reaction containing a high concentration of acid, a collision between acid particles and magnesium atoms is very likely.
second test tube also add 6 mL of 0.1M HCl. Make a solution of 0.165
6. I then rinsed out the beaker and glass rod into the flask to make
of Copper Sulphate. To do this I plan to work out the amount of water
The purpose of this experiment is to use our knowledge from previous experiments to determine the exact concentration of a 0.1M sodium hydroxide solution by titration (Lab Guide pg.141).