- #1
_Bd_
- 109
- 0
Homework Statement
You are titrating 30.0 mL of .3 M CH3NH2 with .15 M HNO3 Kb = 4.4 x 10^-4. Calculate
pH:
-initially
-after 10.0 mL of acid has been added
at the half equivalence point
at the equivalence point
when 70 mL of acid has been added.
Homework Equations
not sure. . .
The Attempt at a Solution
I was thinking that since Kb is given I can get the pOH using the fact that
[OH] = [tex]\sqrt{K_b x Molarity}[/tex] <- im not sure if this is correct tho.
then do an ICE table for the reaction. . . I am not sure about the reaction tho. . .
HNO3 + CH3NH2 >> CH3NH3+ + NO3- . . .is that it?
the first ICE table should be a limiting reagent problem?
knowing the moles of both after 10 mL yields:
HNO3 + CH3NH2 >> CH3NH3+ + NO3-
.0015 | .009 ...| 0....|0 <<<<<<<<<<Initial
-.0015 |-.0015 ...| +.0015 ...| +.0015 <<<<<<<<Consumed
0 ...|.0075 .....| .0015 ...| .0015 <<<<<<< Final
so I can get K at this point right? (K = products/reactants) but that's Ka or Kb?