Background Potassium hydrogen phthalate (KHP for short) is the potassium salt of

Question

Background Potassium hydrogen phthalate (KHP for short) is the potassium salt of the first conjugate base of phthalic acid Since phthalic acid is a diprotic acid, KHP is a monoprotic acid. KHP is used as a primary standard for determining the molarity of base solutions since it is quite stable and easy to handle HCsH404-(aq) + OH-(aq) C8H4042-(aq) + H20(1) +OH O- o- In this practical exam, you will be given a weighed sample of a solid that contains a mixture of KHP and another salt that does not affect the pH of the solution. You will use a sodium hydroxide titration to determine the percent KHP in your sample. If you are red/green colorblind, make sure your laboratory instructor knows before you begin this procedure (several days before you begin). Thymolphthalein indicator will be used instead of phenolphthalein. Thymolphthalein will turn blue at the equivalence point (end point) of the titration. continued on the other side of this pagem

Explanation / Answer

Firstly ,you need to divide your sample into three -four equal parts by weighing.Let the mass of each part be x.Dissolve each part in a fixed volume (say 25 ml) of distilled water in an erlenmeyer flask to make sample solutions of equal concentrations approximately.Add a few drops of phenolpthalein indicator.It gives pink color at the end point for this titration done against standard NaOH solution.Titrate all the sample solutions as trials 1,2,3,4

Let the molarity of NaOH used for titration be =M1

volume of NaOH to reach the end point=V1

Also,volume of sample solution=V2

molarity of sample solution=M2

using equation,M1*V1=M2*V2

M2=M1V1/V2

Once M2 is calculated out ,you can calculate the mol of actual KHP present in the sample solution

mol of actual KHP present in the sample solution=M2*V2 or (=M2*25ml)

now mass of actual KHP present in the sample solution=mol of KHP*molar mass of KHP=m

Now weight % of KHP in the sample=[mass of KHP(m)/sample mass(x)]*100

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