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Report (any type)/Brief report
22/09/2021
In a back titration, you add an excess of standard titrant to the analyte, and then you titrate the excess titrant to determine how much is in excess.
Here’s how you do the calculations.
PROBLEM:
A student added 50.00 mL of 0.1000 mol/L ##”HCl”## to 25.00 mL of a commercial ammonia-based cleaner.
It took 21.50 mL of 0.1000 mol/L ##”NaOH”## to neutralize the excess ##”HCl”##.
What was the concentration of ammonia in the cleaner?
Solution:
##”Part 1. HCl calculations”##
(a) Calculate the moles of ##”HCl”## added to the cleaning solution
##”Moles of HCl” = 0.05000 cancel(“L HCl”) × “0.1000 mol HCl”/(1 cancel(“L HCl”)) = “0.005 000 mol HCl”##
(b) Calculate the moles of ##”NaOH”## used
##”Moles of NaOH” = 0.021 50 cancel(“L NaOH”) × “0.1000 mol NaOH”/(1 cancel(“L NaOH”)) = “0.002 150 mol NaOH”##
(c) Calculate the moles of excess ##”HCl”##
##”HCl + NaOH” → “NaCl + H”_2″O”##
##”Moles of HCl” = 0.002 150 cancel(“mol NaOH”) × “1 mol HCl”/(1 cancel(“mol NaOH”)) = “0.002 150 mol HCl”##
(d) Calculate the moles of ##”HCl”## that reacted with the ##”NH”_3##
##”Moles of HCl reacted = 0.005 000 mol – 0.002 150 mol = 0.002 850 mol”##
##”Part 2. NH”_3 color(white)(l)”calculations”##
(a) Calculate the moles of ##”NH”_3##
##”NH”_3 + “HCl” → “NH”_4″Cl”##
##”Moles of NH”_3 = 0.002 850 cancel(“mol HCl”) × (“1 mol NH”_3)/(1 cancel(“mol HCl”)) = “0.002 850 mol NH”_3##
(b) Calculate the of the ##”NH”_3##
##”Molarity” = “0.002 850 mol”https://studydaddy.com/”0.02500 L” = “0.1140 mol/L”##
I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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