What is the pH at different points in a titration of CH3NH2 with HNO3?

In summary, the conversation discusses the process of titrating .3 M CH3NH2 with .15 M HNO3 using the given Kb value of 4.4 x 10^-4. The pH is calculated at various points, including initially, after adding 10.0 mL of acid, at the half equivalence point, at the equivalence point, and when 70 mL of acid has been added. The suggested method involves using the Kb value to find the pOH and then using an ICE table to calculate the pH. The ICE table includes the reaction HNO3 + CH3NH2 >> CH3NH3+ + NO3- and the calculation involves finding the moles of each component at different points in the tit
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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?
 
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