How can I make back titration calculations?

1 Answer
Apr 17, 2015

Answer:

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.

Explanation:

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 molarity of the #"NH"_3#

#"Molarity" = "0.002 850 mol"/"0.02500 L" = "0.1140 mol/L"#