Problem: Consider the titration of 30.0 mL of 0.050 M  NH3 with 0.025 M  HCl. Calculate the pH after the following volumes of titrant have been added.20.0 mL

🤓 Based on our data, we think this question is relevant for Professor Co's class at OSU.

FREE Expert Solution

We’re being asked to calculate the pH of a solution after the addition of 20.0 mL 0.0250 M HCl solution. The solution is made up of 30.0 mL of 0.050 M ammonia (NH3).


HCl (strong acid) will react with NH3 (base).

▪ NH3 is a base and based on Bronsted-Lowry definition it is a proton acceptor.
▪ HCl is an acid and based on Bronsted-Lowry definition it is a proton donor.

Reaction:

NH3(aq) + HCl(aq) → NH4+(aq) + Cl-(aq)


We will calculate the pH of the solution using the following steps:

View Complete Written Solution
Problem Details

Consider the titration of 30.0 mL of 0.050 M  NH3 with 0.025 M  HCl. Calculate the pH after the following volumes of titrant have been added.

20.0 mL

Learn this topic by watchingWeak Base Strong Acid Titrations Concept Videos
All Chemistry Practice Problems Weak Base Strong Acid Titrations Practice Problems
Q.How do you calculate the pH at the equivalence point for the titration of .190M methylamine with .190M HCl? The Kb of methylamine is 5.0 x 10-4.
Q.A 20.0 mL sample of 0.150 M ethylamine is titrated with 0.0981 M HCl. What is the pH after the addition of 5.0 mL of HCl? For ethylamine, pKb= 3.25.
Q.A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCl.Calculate the pH at 20 mL of added acid.
Q.A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCl.Calculate the pH at equivalence point.
See all problems in Weak Base Strong Acid Titrations