- #1
Mag
consider the titration of 50.0mL of 0.10 M methyl ammonia with 0.10 M HCl. Calculate the pH after
15mL of titrant= 11.05
25mL of titrant= 10.68
50mL of titrant= 8.98
60mL of titrant= ?
At 60 mL there is an excess of 1 mmol of HCl. At this point do I say that the pH is 1 (-log of 0.10)? I'm at a loss.
Mag
15mL of titrant= 11.05
25mL of titrant= 10.68
50mL of titrant= 8.98
60mL of titrant= ?
At 60 mL there is an excess of 1 mmol of HCl. At this point do I say that the pH is 1 (-log of 0.10)? I'm at a loss.
Mag