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"
