# How can I make back titration calculations?

Apr 17, 2015

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 $\text{HCl}$ to 25.00 mL of a commercial ammonia-based cleaner.

It took 21.50 mL of 0.1000 mol/L $\text{NaOH}$ to neutralize the excess $\text{HCl}$.

What was the concentration of ammonia in the cleaner?

Solution:

$\text{Part 1. HCl calculations}$

(a) Calculate the moles of $\text{HCl}$ added to the cleaning solution

$\text{Moles of HCl" = 0.05000 cancel("L HCl") × "0.1000 mol HCl"/(1 cancel("L HCl")) = "0.005 000 mol HCl}$

(b) Calculate the moles of $\text{NaOH}$ used

$\text{Moles of NaOH" = 0.021 50 cancel("L NaOH") × "0.1000 mol NaOH"/(1 cancel("L NaOH")) = "0.002 150 mol NaOH}$

(c) Calculate the moles of excess $\text{HCl}$

$\text{HCl + NaOH" → "NaCl + H"_2"O}$

$\text{Moles of HCl" = 0.002 150 cancel("mol NaOH") × "1 mol HCl"/(1 cancel("mol NaOH")) = "0.002 150 mol HCl}$

(d) Calculate the moles of $\text{HCl}$ that reacted with the ${\text{NH}}_{3}$

$\text{Moles of HCl reacted = 0.005 000 mol – 0.002 150 mol = 0.002 850 mol}$

$\text{Part 2. NH"_3 color(white)(l)"calculations}$

(a) Calculate the moles of ${\text{NH}}_{3}$

$\text{NH"_3 + "HCl" → "NH"_4"Cl}$

${\text{Moles of NH"_3 = 0.002 850 cancel("mol HCl") × ("1 mol NH"_3)/(1 cancel("mol HCl")) = "0.002 850 mol NH}}_{3}$

(b) Calculate the molarity of the ${\text{NH}}_{3}$

$\text{Molarity" = "0.002 850 mol"/"0.02500 L" = "0.1140 mol/L}$