DATA SHEET 4.3: Standardization of a solution using potassium hydrogen phthalate

Question

DATA SHEET 4.3: Standardization of a solution using potassium hydrogen phthalate Balanced reaction- k HaMat NaOH tlanat NaKC4Hwa Trial #3 (if necessary) Titration Trial #1 Trial #2 Mass of clean and empty Erlenmeyer flask | (27.39%) 132.2927 Mass of Erlenmeyer flask with KHP oszt5-0A45- Measured KHP mass 2 Calculated KHoles 00.00244 Initial buret reading | One | 60m Final buret reading | Izq39md2.gvnL A57m2 3 Volume of titrant added | Za59nL| 2St3- 4 Moles of titrant added Calculated molarity sodium hydroxide Average molarity, sodium hydroxide

Explanation / Answer

Given calculated moles of KHP =0.002569 and titrant volume = 29.38 mL i.e. convert into L (0.02938 L)

mole of NaOH is equal to moles of KHP at the end point of reaction.

calculated moles of KHP = number of moles of NaOH

For Trail 1

number of moles of NaOH=0.002569 (mole of KHP )

molarity = num of moles / volume in litre =0.002569 /0.02938 = 0.08744 M

molarity = 0.08744 M (Molarity for trial 1 say it as M1)

similarly For trail 2

Number of moles of NaOH = 0.002441 moles volume of titrant =28.43 mL i.e. convert into L ( 0.02843 L)

Molarity of NaOH = 0.002441/0.02843 =0.08586

Molarity of NaOH = 0.08586 M (molarity For trail 2 say it as M2)

Now calculate average molarity of NaOH

Average molarity of NaOH = (M1 + M2 ) / 2 = (0.08744 + 0.08586 ) / 2 =0.08665 M

Average molarity of NaOH =0.08665 M

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