• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

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

Extracts from this document...


Joshua Dawber BTEC Level 3: Unit 1 Fundamentals of science Mr Nolan Making a Standard Solution and Titration to Find Molarity of Unknown Acid 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. ...read more.


23+16+1=40 this is where the 40 comes from.) 0.99 ÷ 40= 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 ...read more.


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. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our AS and A Level Physical Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related AS and A Level Physical Chemistry essays

  1. Marked by a teacher

    Finding out how much acid there is in a solution

    Acids Vs Bases (titration graphs) Acid/Base titration helps to present the titration of acids and bases. The PH lies on the y axis and the titrant volume lies on the x-axis. Titration graphs helps to outline the unknown solution. Strong Acid Vs Strong Base e.g.

  2. Preparation of Standard solution and Standardization of Hydrochloric acid

    Fourth, diluted titrant and diluted titrated solution may be accidentally used in the titration if the burette or pipette are not rinsed with solution to be contained or transferred after being rinsed with distilled water. Consequently, titrant is slightly diluted.

  1. Acid-Base Titrations.

    approximately equal to [CH3COO-], from the stoichiometric dissociation of the weak acetic acid and the fact that it is the major source of hydrated protons in the solution. Solving: [H3O+]2 = 1.75 x 10-5 x 0.1, [H3O+] = 1.32 x 10-3, pH = 2.88 Before the Equivalence Point The pH

  2. Lab Report. Objective: To determine the concentration of unknown standard sodium hydroxide solution ...

    As a result, the calculated enthalpy change of neutralization would be lower than the actual one. To reduce this error, a Styrofoam cup instead of a polystyrene cup can be used since Styrofoam is a better insular of heat than polystyrene.

  1. Identification of an Organic Unknown

    To do this I will do a simple experiment by adding a few drops of universal indicator to a test tube with the unknown organic compound. This will identify to me if there is an alcohol functional group present. If this functional group is present then the solution would go

  2. Investigating the Rate of the Reaction between Bromide and Bromate Ions in Acid Solution

    This, with the Arrhenius Equation, can be used to find the activation enthalpy of the reaction (see above - page 12). I will then repeat the reaction using the same set concentrations, but adding ten drops of a transition metal ion solution, which could work as a catalyst.

  1. Investigating how concentration affects rate of reaction

    M1 For example, to make a 250ml solution with concentration 0.008M of potassium bromide from a solution of 0.01M, you would do the following: V1 = (0.008 x 250ml) 0.01 = 200ml This means that you would need to use 200ml of the original 0.01M solution with 50ml of distilled water to make 250ml of 0.008M solution.

  2. This investigation is an investigation to find the concentration of two unknown solutions, Na0H ...

    Hydroxide took to neutralise 25cm3 of Sodium Hydrogen Sulphate (ml)±0.05 1 21.0 2 20.5 3 20.2 4 21.0 5 21.0 6 20.8 7 20.8 8 20.7 9 19.7 10 19.3 Average 23.3 Standard Deviation 23.3±0.6 Discussion 1st Titration By looking at the result it could be argued that the first result (21.8ml)

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work