Acid-Base Titration Report Sheet A. Acetic Acid in Vinegar A.1 Brand, A2 Molarit

Question

Acid-Base Titration Report Sheet A. Acetic Acid in Vinegar A.1 Brand, A2 Molarity (M) of NaOH (stated on label Volume S0mL Volume samL (% on label) Trial I Trial 2 Trial 3 A3 Initial NaOH level A4 Final NaOH level A.5 Volume (mL) of in buret in buret NaOH used Average volume A.6 Average volume in liters (L.) A.7 Moles of NaOH used in titration mole NaOH Show caleulations) A.8 Moles of HC, H,O, neutralized by NaOH A.9 Molarity of HC H,0 M HC, H,0 Show calculatiorns A.10 Grams HC H 02 g HC HyO2 Show caleulations A.11 Percent (m/v) HC H,O, in vinegar -% HC,H,02 Show calculations Questions and Problems Q.1 How many grams of MgrOH), will be needed to neutralize 25 mL of stomach acid if stomach acid is 0.10 M HCI? Q2 How many mL of a 0.10 M NaOH solution are needed to neutralize 15 ml. of 0.20 M H,PO solution? 115

Explanation / Answer

All the data to be filled in table should be coming from lab itself

1.

Lets write the reaction:

Mg(OH)2 + 2HCl ------> MgCl2 + 2H2O

1 mol magnesium hydroxide needs 2 mol HCl

moles of HCl in stomach = molarity x volume in litres = 0.1M x 0.025L = 0.0025 moles

moles of magnesium hydroxide needed = 0.0025 moles / 2 = 0.00125 moles

molar mass of Mg(OH)2 = 58.3197 g/mol

mass = moles x molar mass = 0.00125 moles x 58.3197 g/mol = 0.0729 g

2.

3NaOH + H3PO4 ------>Na3PO4 + 3H2O

moles of H3PO4 = molarity x volume in litres = 0.20M x 0.015L = 0.003 moles

moles of NaOH needed = 3 x moles of H3PO4  {as can be seen from reaction equation}

= 3 x 0.003 moles = 0.009 moles NaOH needed

moles = molarity x volume = 0.009 moles = 0.1M x v

v = 0.09L = 90 mL

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