Hi, please help me complete trial 1, step-by-step. The lab is Acid-Base Titratio

Question

Hi, please help me complete trial 1, step-by-step. The lab is Acid-Base Titrations.

We mixed roughly 1 g of KHP and 10 mL water in a flask. Added drops of phenolphthalein indicator. Then titrated NaOH into flask containing KHP and H2O until it reached equivalence point.

Thank you

Part I. Standardization of NaOH solution: Trial #2 Trial #1 Measurements 0.470 g Mass of weighing paper 1.480 g Mass of weighing paper KHP Mass of KHP 1.01 g of moles of KHP 43 mL Initial buret reading 31 mL Final buret reading Volume of NaOH 12 mL of moles of NaOH Concentration of N Calculations KHP -204.44 g/mol

Explanation / Answer

Given

Mass of KHP = 1.01 g

Molar mass of KHP = 204.44 g/mol

No. of moles of KHP = Mass of KHP / molar mass of KHP = 1.01 g / 204.44 g/mol = 4.94 * 10-3 mole of KHP

reaction between KHP and NaOH is

KHC8H4O4 + NaOH = NaKC8H4O4 + H2O

1 mole 1mole 1mole 1 mole

so each mole of KHP require 1 mole of NaOH

so No. of moles of NaOH = 4.94 * 10-3 mole Answer

Volume of NaOH = 12 ml = 0.012 L

Molarity = No. of moles / Volume = 4.94 * 10-3 mole / 0.012 L = 0.412 mol/L

Concentration of NaOH = 0.412 M Answer

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