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# Making a Standard Solution and Titration to Find Molarity of Unknown Acid Essay Sample

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• Word count: 912
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• Category: acid

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## Introduction of TOPIC

To find the molarity of the unknown acid, first we had to create a standard solution, the solution we created was Sodium Hydroxide (NaOH). We wanted a 0.1 molar solution of sodium hydroxide so to get this we had to dissolve 4g of NaOH into 1000cm³ of water, but we didn’t want 1000cm³ we wanted 250cm³ so to work out how much sodium hydroxide would be needed you need to do the same equation to the number of grams (g) than with the volume of water, so to get 1000cm³ down to 250cm³ you divide it by 4, so you divide 4 by 4 which gives you 1, so one gram of NaOH is needed to make a 0.1 molar solution in 250cm³ of water. Next is making the solution, the equipment needed to make this standard solution is: a balance, beaker, volumetric flask, glass rod, wash bottle. And the ingredients for the solution are NaOH and distilled water.

To make NaOH solution is to measure out 1g of sodium hydroxide and place on a scrap piece of paper which is on the balance, it isn’t essential that you get exactly 1g just approximately 1g. Then put some distilled water into a beaker enough to dissolve the sodium hydroxide, transfer the sodium hydroxide from the paper to the beaker and dissolve by swirling and stirring. Once dissolved transfer this solution to a volumetric flask, and wash out the beaker and glass rod which was used to stir the solid NaOH into the water, now add distilled water to the volumetric flask, up until the bottom of the meniscus is on the 250cm³ line and shake and mix it up a little, then you have made your solution.

In my solution it wasn’t 1g, I weighed 0.99g. The next stage is to calculate the molarity of your solution. To work out the moles it is moles= grams ÷ relative molecular mass (RMM) so for my solution it will be 0.99÷40 (40 is the RMM of sodium hydroxide, this is calculated by adding the mass of each atom in the compound toge

ther, so for NaOH it is Na=23 O=16 and H=1. 23+16+1=40 this is where the 40 comes from.) 0.99 ÷ 40=

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0.02475 rounded to 4 decimal places is 0.0248 that is the molarity of the 250cm³ but molarity is always measured in 1000cm³ so now you have to times 0.0248 by 4, 0.0248 x 4= 0.992, and that is the final molarity of your solution so my molarity is 0.992M.

Now is to titrate you solution with the unknown acid, to do this you need: a clamp, a beaker for acid, a beaker for your standard solution and another beaker for waste, a conical flask, 50ml burette, 25ml pipette. Once all the equipment has been set up you now need to add your unknown solution into the burette and leave the tap open and put the waste beaker under it to make sure there is no air bubbles in the burette, turn the tap off and fill the burette up, now take the pipette filler and fill up your pipette with your standard solution and put that in the conical flask, add a colour indicator to the conical flask and put the conical flask under the burette open the tap, and you are looking for the first colour change that lasts for approximately 10 seconds, repeat the titration until you have 3 results within .1 of each other. In my titrations I did 4, the first result was 22.6ml used, the second was 23.1ml, third was 22.7ml and the final one was 22.8ml. Now the calculation for the molarity of the acid can be solved.

The first step in working out the concentration of the unknown acid is balancing the equation. The equation for our experiment is: NaOH + HCl ï NaCl + H2O and this equation is already balanced because there is 1 atom of Na on each side, 1 atom of O on each side, 2 atoms of H on either side and 1 atom of Cl on each side. So this reaction is a 1:1 reaction. The reasons this is a 1:1 reaction can be found in the periodic table, the RMM of each side of the equation has to be the same and to work this out you need the atomic mass, Na=23, O=16, H=1 (x2) and Cl=35. The atomic mass is the larger of the two numbers on the periodic table found with an element. The total of these atomic masses is 76. And it is exactly the same on the other side it is just that the compounds are different, this is due to the groups on the periodic table that they are in and that determines the bonds between atoms.

The equation to work out the concentration of the unknown acid is: moles x 1000 ÷ average titration. The average titration is all the titration results added together and divided by 4, but we are going to discard the 23.1ml result because it isn’t close enough to the other three so is recognised as an anomaly, so (22.6 + 22.7 +22.8)÷3 = 22.7cm³ so now using the equation you can work out the concentration of the acid. (0.0248 x 1000)÷ 22.7 = 0.1093, the actual concentration of the acid was 0.0984. My predicted concentration is 0.0109 above the actual concentration this could be due to inaccuracies with the measuring of the mass of NaOH to begin with also wrongly measuring the amount of my standard solution was used to titrate the acid.

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