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.
We then took 1ml of the 0.1% solution from test tube 2 using the glucose pipette and added it to test tube 3, we then used the H2O pipette and added 9ml of H2O into test tube 3 creating 10ml of 0.01% solution.
Once the mixture had been completely dissolved, the solution was transferred to a separatory funnel. The solution was then extracted twice using 5.0 mL of 1 M
Put the amount of 0.1M cobalt (II) chloride hexahydrate that fills the end of a spatula into a test tube. Then add 2mL of 95% ethanol. Tap the end of the test tube to mix the solution and record the pertinent data in section 1 of the Data Table. Discard the solution in the appropriate container as directed to you by your lab instructor.
We were then to make a base solution of 0.7 M NaOH. In order to standardize
Apparatus: * 1 measuring cylinder * 1 test tube * 1 stop clock * A large gelatine cube containing indicator and NaOH * Hydrochloric acid ranging from 1-3 molars * A scalpel Diagram: Method: * Take the large gelatine cube and cut into 15 equal pieces * Place on piece of the cube into the test tube * Measure out 10mls of HCl in the measuring cylinder * Pour the HCl into the test tube with the gelatine cube and start the clock * Time how long it takes for the pink colour inside the gelatine cube to completely disappear * You will also notice that the cube dissolves slightly * Record your results and repeat this same process 3 times for each molar of acid: § 1 molar § 1.5 molar § 2 molar
The amount of hydrochloric acid. 3. The concentration of the hydrochloric acid. 4. The surface area of the calcium carbonate.
= 3 ´ E(C-H) + 1 ´ E(C-O) + 1 ´ E(O-H) + 1.5 ´ E(O=O)
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
Reaction 2: H = 50 x 4.18 x -10.3" H = -2152.7 This value is for 1.37g of calcium oxide, not 56.1g, which is its relative molecular mass. Therefore: H =
Determining the Concentration Of Limewater Solution Aim: The aim of this experiment is it to find out the concentration of Limewater by performing a titration with hydrochloric acid which has concentration exactly 2.00M.. What is required for me is that I have to design my own experiment and chose the right and appropriate apparatus and equipment. I will be provided with 250cm3 of limewater, which has been made to which contains approximately 1g/dm3 of calcium Hydroxide. This hypothesis from www.studentcentral.co.uk We were also give Hydrochloric acid (HCl) with a concentration of 2.00 mol/dm3 normal laboratory apparatus was also given and so was an indicator.
of Copper Sulphate. To do this I plan to work out the amount of water
second test tube also add 6 mL of 0.1M HCl. Make a solution of 0.165
In this experiment three different equations were used and they are the Stoichiometry of Titration Reaction, Converting mL to L, and Calculating the Molarity of NaOH and HCl (Lab Guide pg. 142 and 143).
Weigh accurately about 0.3 g and dissolve in 40 ml of me-thanol. Titrate with 0.1 M so-dium hydroxide.