posted by some kid on .
In the titration of 50.0 mL of 1.0 M CH3NH2 (kb=4.4 x 10^-4), with 0.50 M HCl, calculate the pH
a) after 50.0 mL of 0.50 M has been added
b) at the stoichiometric point.
At the 50 mL mark, you have the following:
M x L = 1 M x 0.050 L = 0.05 moles CH3NH2 to start.
You have added 0.050 L x 0.50 M = 0.025 moles HCl.
What's left after they react?
You have used 0.025 moles CH3NH2 which leaves 0.025 remaining. You have used all of the HCl. You have formed 0.025 moles of the salt. So you have a buffer consisting of 0.025 moles of the base and 0.025 moles of the salt. Use the Henderson-Hasselbalch equation.
pH = pKa + log(base/acid).
(I suppose you know that pKa = 14-pKb)
At the stoichiometric point, you will have 0.050 moles of the salt. The pH will be determined by the hydrolysis of the salt.
CH3NH3^+ + HOH ==> CH3NH2 + H3O^+
Make an ICE chart and solve for H3O^+ and pH from that.
Kb = (CH3NH2)(H3O^+)/CH3NH3^+)
Check my work.
for the ICE chart, does it need to be in moles or molarity?
I would do it in moles but I don't think it makes any difference.
@somekid: ICE Charts have to be either in molarity or pressures.