Question

Questions:

Acid-Base Titrations continued Standardization Data Table Trial 1 Trial 2 Trial 3 Mass KHP Final Volume, mL Initial Volume, ml Volume of NaOH added, mL C.40 29 mL Molarity NaOH (Average)- G.0220 veght M O G232 Equivalent Mass Data Table perse 232 Trial 1 Trial 2 Mass Acid, g Final Volume, mL Initial Volume, mL Volume of NaOH added, mL 2 ml 0 Equivalent Mass (Average) g/mol aper o.29 pK, Data Table Mass of Unknown Acid 0 833 Standard NaOH Concentration ZC S Initial Buret Reading Initial pH Buret Reading (mL) pH Buret Reading (Con't.) 3 20 hl 21 ml 4 26 . 30 a mL 3 434 H. 31 23 m

A. From the standardization data, calculate the molarity of the sodium hydroxide solution for each trail. Average the values and enter the average in the Standardization Data Table.

B. From the equivalent mass data, calculate the equivalent mass of the unknown acid for each trial. Average the values and enter the average in the Equivalent Mass Data Table.

Answer 1

KHP + NaOH

1 mole KHP is neutralized with 1 mole of NaOH

Standardization of NaOH

Trial 1, moles KHP = 0.406 g/204.22 g/mol = 0.002 moles

moles NaOH needed = 0.002 moles

molarity NaOH = 0.002 moles/0.0205 L = 0.0976 M

Trial 2, moles KHP = 0.407 g/204.22 g/mol = 0.002 moles

moles NaOH needed = 0.002 moles

molarity NaOH = 0.002 moles/0.0205 L = 0.0976 M

Trial 3, moles KHP = 0.409 g/204.22 g/mol = 0.002 moles

moles NaOH needed = 0.002 moles

molarity NaOH = 0.002 moles/0.0205 L = 0.0976 M

average molarity of NaOH = 0.0976 M

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Equivalent mass of acid

Trial 1, moles NaOH used = 0.0976 M x 0.014 L = 0.00137 moles

moles acid present = 0.00137 moles

equivalent mass of acid = 0.496 g/0.00137 moles = 363 g/mol

Trial 2, moles NaOH used = 0.0976 M x 0.012 L = 0.00120 moles

moles acid present = 0.00120 moles

equivalent mass of acid = 0.496 g/0.00120 moles = 413.33 g/mol

Average equivalent mas of acid = 383.165 g/mol