1. Prepared by Chris Weller

2. Laboratory Glassware:

3.  Mole Mass Conversion  Grams/L x (1/MW) = Moles/L Example: Need to know how many grams of NaOH to make 1Litre of 2M. Rearrange equation: to show that the grams is the unknown we what to find. Grams= Moles / (1/MW) Grams =2M / (1/40.01) Grams= 80.02 of NaOH /Litre of distilled water.  Caution this is exothermic reaction.

4.  Important to remember check the label you need to know the starting strength (%) Grams = Density x 1000ml x % Example: HCl comes in two different strengths 32% and 36%. Grams = 1.179g/ml x1000ml x 36% Grams = 1.179g x1000 x0.36 Grams = 424.44

5.  Next step to work out the number of moles. Grams x (1/MW) = Moles  Rearrange to find the number Moles. M= (g x 1000ml) / (MW x ml) M= (424.44g x 1000) /(36.46 /1000) Moles = 11.64M This is the strength of your HCl  Now work out how many Moles is 32% HCl.

6.  Next Step is dilution. Example : 32% HCl = 10.17 M and you need to make 500mls of 2M. M1x V1 = M2 x V2 M1=The original number of moles of HCl 10.17 V1= How much do we need? M2 = Moles needed 2M V2 = Volume needed 500mls 10.17M x V1ml= 2Mx 500ml Rearrange to find V1 V1= (2x500)/10.17 V1= 98.33ml

7.  Sulphuric acid. 96-98% 1. Grams = Density x 1000ml x % 2. M= (g x 1000ml) / (MW x ml) Step1. Grams= 1.84g/mlx 1000mlx 96% Grams=1.84g x 1000ml x0.98 Grams=1803.2 Step 2. M= (1803.2 x1000ml) / (96.07x/1000ml) M= 18.85 Then dilute to suit the moles you require.

8.  When mixing solutes what is the final Mole?  Example: 50ml of 0.5M NaOH with 250ml of 1M NaOH What is the final molar strength. M1V1+M2V2= M3V3 M1= 0.5M ; V1 = 50ml; M2 = 1.0; V2 =250ml M3 = unknown; V3 = 300ml Rearrange to find M3 M3= (0.5Mx50ml+1.0M x 250ml) /300 M3=0.92M

9.  Does the Chemical have water added? This needs to be taken into consideration when considering the Molecular Weight (MW).  Read the information on the Chemical container label. Example. Copper Sulphate comes in Anhydrous CuSO4(pale green to white powder) and Pentahydrate. CuSO4.5H2O (bright blue powder) CuSO4 =159.62g/mol CuSO4. 5H2O = 249.70g/mol This will affect your accuracy of your chemical solution.  Liquids chemicals always check the % and density

10.  NaOH example: 80.02g/L to make 2M  You only need 250mls.  1000/250 = 4  80.02/4 = 20.005g Therefore you need 20.005g of NaOH to make up 250mls.

11. Composition of concentrated reagent grade acids, ammonium hydroxide, and sodium and potassium hydroxide solutions (with dilution directions to prepare 1N solution) ChemicalMFApprox. Strength ofMolarity ofMilliliters of Concd. Reagent Necessary Concd. Reagent a Concd. Reagentto Prepare 1 Liter of 1 Normal Soln. c Acetic Acid, GlacialCH 3 COOH99.817.457.5 Formic AcidHCOOH9023.642.5 Hydrochloric AcidHCl37.212.182.5 Hydrofluoric AcidHF4928.934.5 Nitric AcidHNO370.415.963 Perchloric AcidHClO470.511.785.5 Perchloric AcidHClO461.39.5105.5 Phosphoric AcidH 3 PO485.514.822.5 Sulfuric AcidH 2 SO4961828 Ammonium HydroxideNH 4 OH56.6 b 14.569 Sodium HydroxideNaOH50.519.451.5 Potassium HydroxideKOH4511.785.5 a - Representative value, w/w%. b - Equivalent to 28.0% w/w NH 3. c - Rounded to nearest 0.5 ml.

12.  Example Nitrate testing of water samples. Students need a more accurate result to compare samples as all the sample appear to be the same reading. So you make up the standards with in the range they ask for. E.g. 0-10 ppm then the Student can use Beer-Lambert Law to make a calibration curve.

13.  0-10ppm = 0-10mg/L  Can use Potassium or Calcium nitrate. To make up the solutions. As K and Ca are found in the water normal so the metals won’t affect the results.  Consider : how accurate are you scales? How many digital placing? This will affect how you approach making the standards. For this exercise we will assume the scales read two decimal places.

14.  Accuracy is important. Need to calculate the standards require. A range between 0 and 10 mg/L. The more samples you have the better the accuracy. The norm is between 5-10 samples for this range including your blank sample.  0, 1,2,3,4,5,6,7,8,9,10 mg/L  0,1,2.5,5,10mg/L  If you don’t want to make up a Litre you can scale it back to 100mls.

15.  10mg/L equals 0.01grams in 1000mls. Our problem is that our scale only reads 2 decimal places. What are our options. 1. Go and buy better scales $$ 2. We could do a series of dilutions. Series Dilutions: Step 1. Make up 0.01g of CaNO3 in 1000mls of water. This is our stock solution. You may have to may up the fresh standards daily as this is a colorimetric method. The colour can either fade or intensify, depends on what it is.

16.  Step 2. Once you have thoroughly mixed the stock solution. Decant 100mls in to a volumetric flask. Mark it as 10mg/L. of CaNO3  Step 3. Into another 100ml volumetric flask pipette 50 mls of stock solution and then fill with distilled water to the mark. This is 5mg/L CaNO3  Step 4. Pipette 50mls of the 5mg CaNO3 into a 100mls volumetric flask then fill with distilled water to the mark. This is 2.5mg/L CaNO3.

17.  Step 5. Pipette 10mls of the 10mg standard into a 100ml volumetric flask and fill to the mark. This is 1mg/L of CaNO3.  Step 6. Pipette 5 ml of your 1 st standard 10mg then add the drops of indicators as per the instructions in the kit. Wait the required time and them place in cuvette and then read the sample in your colorimeter.  Repeat for all other standards.  Then using excel you can plot your Standard Curve Chart. Then test the unknown.

18.  This method can be done for all colorimetric and UV Vis analysis.  For use in EEI the students can either make the standards up themselves and compare their standard curve results against yours.  Or the students can use your standard curve results and then insert their own data to give them a more accurate reading in ppm.

19.  Oriakhi, Christopher O. Chemistry in Quantiative Lanugage: Fundamentals of General Chemistry.