[mike_302]

BIIIIG question

Calculate the pH of a buffered solution made up of 0.015 M Sodium acetate and 1 L of 0.10 M acetic acid.

I, unfortunatly, have no idea where to even START on this question :S

please *delete me*

[Borek]

http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch

[mike_302]

Okay, okay.

Here's the real man's question

50mL of .2 M NH3 is irated with .2M HCl . The Kb for NH3 is 1.8*10^-6

b) Added 10.00mL HCl

part 1, neutralization
take concentrations and volumes, find number of mols, and subtract mols of HCl from mols of NH3 to find the remaining mols of NH3  which works out to 8*10^-3 mols   

Work that back to concentration (using the NEW total volume of .060L) -->   .1333M of NH3

part 2, equilibrium
so I thought there should be .1333M NH3 (since that was all that remains after the neutalization) , 0M of OH- ions, and   NH4, I thought , should be the same as there was HCl, since HCl and the consumed NH3 is what formed the NH4 . so I am thinking there is .03333M NH4 to start

NH3   <--->   OH-    +    NH4+
.1333M         0               .0333
-x                +x              +x
.1333-x         +x             .0333+x


So, following through with the Kb calculations and whatnot, I don't get the same pH as the teacher's answer of 8.26

Yes, I use Kb to get OH, then subtract that from 14 to get pH..

doesn't work

[AWK]

The Kb for NH3 is 1.8*10^-6

[Borek]

part 2, equilibrium
so I thought there should be .1333M NH3 (since that was all that remains after the neutalization)

What is NH₄⁺ concentration?

You have a solution of both base and conjugated acid. How do you call such solution? How do you calculate its pH?