CHNH2 + HCl ==> CH3NH3Cl
b. Look up the delta Go values in your text/notes and solve for dGo for the rxn, then dGo = -RTlnK. Substitute and solve for K.
c.
..CH3NH2 + HOH ==> CH3NH3^+ + OH^-
I..0.030............0..........0
C.....-x............x..........x
E..0.030-x..........x..........x
Convert pKb to Kb (pKb = -log Kb). Substitute the E line into the Kb expression and solve for x = OH^- and convert to H^+ then to pH.
d.
millimols CH3NH2 initially = mL x M = 30 mL x 0.030 M = 0.9
Therefore, CH3NH3Cl formed is 10 mL x 0.025 = 0.25 mmols and the CHNH2 used is 0.25 which leaves 0.9-0.25 = 0.65. Substitute the 0.65 and the 0.25 into the Henderson-Hasselbalch equation and solve for pH. You will need pKa for that and you get it from pKa + pKb = pKw = 14. With pKw and pKb, you can solve for pKa.
Methyl amine is a weak base with a pKb=3.35. Consider the titration of 30.0mL of .030M of Methyl amine with 0.025M HCl.
a) Write the appropriate equation for the reaction.
b) Calculate K for the reaction in part (a).
c) Calculate pH of the initial methyl amine solution (0mL added HCl)
d) calculate pH of solution with addition of 10mL HCL
1 answer