A 75.0-mL volume of 0.200 M NH3 (Kb=1.8 imes 10^{-5}) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 28.0mL of HNO_3.

Question

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8	imes 10^{-5}) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 28.0mL of  HNO_3.

vincent-lyon.fr · Answer

What is the reaction that is taking place?

anonymous · Answer

You are going to use the Henderson law (you can use this when the kb value is less than -5)
first solve for pkb which is -log [kb]
solve for concentration of NH3
then to solve for pH
pH= pka + log [salt]/[base]
determine whether a foreign base or acid is being added to the pH.
If it's an acid, you are going to add that foreign acid, and minus from the base.
If it's the base, you add to the base, and minus from the acid.
Once you solve for the log [salt]/[base] if the number is negative, you minus that number from your pka, if it's positive you add to your pka to determine your pH.
I hope this helps!