CALCULATIONS A. NaOH Preparation 0.4 M Approximate molarity of NaoH B. NaOH Stan
ID: 514831 • Letter: C
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CALCULATIONS A. NaOH Preparation 0.4 M Approximate molarity of NaoH B. NaOH Standardization: Trial 032g 1.9359 1.1 Mass of KHP Volume of NaOH Molarity of NaOH Average ty of 0, 13 30 m molari NaOH C. Molar Mass of NHX Salt I. Volume of NaOH 00 m LS oon 25.00 mL 2. Molarity of NaOH 3. mol NaOH added 4. Volume of HCO 13 Ile-goreu is 10m 5. Molarity of H CO, 6. mol H2Co, used 7. mol NaOH reacted with NH X 8. mol NH X in sample U20010 Wul 9. Mass of sample 10. NH X molar mass Average NHrx molar mass:Explanation / Answer
Moles of H2C2O4 = Molarity of H2C2O4 * volume of H2C2O4 in L
Moles of H2C2O4 for trial 1: 0.25 M * 0.0136L = 3.4*10^-3 moles
trial2 : 0.25 M * 0.0169 L = 4.23*10^-3
trial 3 : 0.25 M * 0.0159 L = 3.98*10^-3
The reaction of H2C2O4 with NaOH is : H2C2O4 + 2NaOH -----> Na2C2O4 + H2O
1 mol H2C2O4 reacts with 2 moles NaOH. So, amount of naOH that has reacted with H2C2O4 is :
trial 1: 2 * 3.4*10^-3 moles =6.8*10^-3 moles
trial2 : 2* 4.23*10^-3 = 8.46 *10^-3 moles
trial 3 : 2* 3.98*10^-3 = 7.96*10^-3 moles
Amount of NaOH initially present = 0.433 M * 0.025 = 0.0108 moles
NaOH at first will react with NH4X. After the reaction is complete, excess NaOH will react with H2C2O4.
Amount of NaOH that has reacted with NH4X is :
Trial 1 : 0.0108 - 6.8*10^-3 = 0.004 moles
Tria 2 : 0.0108 - 8.46 *10^-3 moles =0.00234 moles
Trial 3 : 0.0108 - 7.96*10^-3 moles = 0.00284 moles
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