Question

Find the pH of a buffer solution that is 0.90 M NH3 and 0.80 M in NH4Cl. Find the pH of the solution after 0.10 mol of HCl has been added to 1.00 L of the solution.

Please show clear/concise step by step instructions, thank you!

Answer 1

To calculate the Ph we use Henderson-Hasselbalch equation as follows:

pH = -log Ka + log ([NH3/NH4Cl])

Ka = (1.0x10^-14/Kb)

The Kb of NH3 is 1.80x10-5

Ka = 10^-14/ (1.8 * 10^-5 ) = 5.56 * 10^-10

pH = -log 5.56 * 10^-10 + log ( 0.90/0.80])

= 9.25 + log 1.125

= 9.25 + 0.0512

= 9.30

b) When you add HCl, it will react with NH3 in the buffer.
                 NH3 + HCl --> NH4Cl
Before: 0.90 mole 0.10 mol 0.80 mol
Change: -0.10 mol -0.10 mol +0.10 mol
After: 0.80 mol 0.00 mol 0.90 mol
Now [NH3] = 0.80 M and [NH4Cl] = 0.90 M

therefore ;

pH = -log 5.56 * 10^-10 + log ( 0.80/0.90])

= 9.25 + log 0.888

= 9.25 - 0.0512

= 9.20