Calaculate the %ammonia (by mass) in Nickelammine complex Calculate the mmoles o
ID: 553130 • Letter: C
Question
Calaculate the %ammonia (by mass) in Nickelammine complex
Calculate the mmoles of NH3/gram of Nickelammine complex
Mass of crystalline nickel ammine complex = 0.2154 g
Vol NaOH used = 20.05 mL
Additional information that may be of use (Lab procedure)
1. Weigh accurately by difference, three 0.2 g samples of the violet crystalline nickel ammine complex into three 250 kml erlenmyer flasks
2. To each erlenmyer flask pipet 15.00 ml of 0.4999M HCl solution
3. Swirl to dissolve samples
4. Add 20 ml distilled water and 10 drops of methyl red indicator to each flask
5. Obtain, clean and rinse buret with 0.1066M NaOH. Titrate each of the three solutions with NaOH to a colour change from pink to yellow
*****If a student unknowingly overtitrated the methyl red endpoint in the ammonia determination, how would that affect his/her NH3/Ni2+ ratio? ******
Explanation / Answer
Ans:
Theory:
A titration is done to determine the percentage of Ammonia in the complex.
This is done through an acid-base back-titration.
Ammonia from the complex is liberated by dissolution in water, then reacted with excess HCl to bind all the ammonia as NH4Cl, then excess acid is determined by acid-base titration.
Reactions Used:
[Ni(NH3)6] + 6H2O -> [Ni(H2O)6] + 6NH3
HCl (aq) + NH3 (aq) -> NH4Cl (aq)
HCl (aq) + NaOH (aq) -> NaOH (aq) + H2O
Calculation:
Mass of crystalline nickel ammine complex = 0.2154 g
Vol NaOH used = 20.05 mL
Molarity of NaOH used = 0.1066M
moles of NaOH = Molarity x Litres = 0.1066 x 0.0205 = 0.0021853 moles NaOH
Since 1 mole of NaOH reacts with 1 mole of HCl,
a. moles of excess HCl in solution [or HCl that reacted with titrant NaOH]
= 0.0021853 moles HCl
b. 15.00 ml of 0.4999M HCl is taken
initial moles of HCl = Molarity x Litres = 0.4999 x 0.015
= 0.0074985 moles HCl taken initially
c. moles of HCl used up = HCl taken initially - excess HCl(found by titrating with NaOH)
= 0.0074985 - 0.0021853
= 0.0053132 moles HCl used up in the reaction between NH3 and HCl.
d. To find the moles of NH3 reacted with HCl:
No. of moles of NH4+ formed = No. of moles of H+ reacted with [Ni(NH3)x]2+ = No. of moles of NH3 present in complex = = 0.0053132 moles HCl = 5.3132 x 10-3 mol
Molar mass of NH3 = 14.01 + 1.01 x 3 = 17.04
Mass of ammonia in [Ni(NH3)x]2+ = 5.3132 × 10-3 × 17.04 = 0.0905369 g
Weight % of ammonia = 0.0905369 / 0.2154 x 100 % = 42.03 %
ans:
I. Weight % of ammonia in the complex = 42.03%
II. If the student overtitrated the methyl red endpoint in the ammonia determination,
i) the moles of NaOH titrated will be more consequently the moles of HCl will be more than the actual amount.
ii) Thus the excess HCl will be more and the HCl used up or reacted with NH3 will be less than the actual amount.
iii) Hence the moles of NH3 that reacted with HCl will also be less than the actual value.
iv) Finally his/her NH3/Ni2+ ratio will be less than the actual value.