Question

Buffer question

Answer 1

i got 4.14 for my answer, why is the answer 4.26 instead?

Answer 2

There's lots of potential mistakes, without seeing your work step by step its impossible to say

Answer 3

\[\left[ HA \right]=1.0M \times \frac{ 0.5L }{ 1.25L}=0.4M\]
\[[A ^{-}]=0.25M \times \frac{ 0.5M }{ 1.25L }=0.1M\]
i used the henderson-hasselbach and got 4.14

Answer 4

pKa= -log(1.8x10^-5) = 4.74

Answer 5

but once the NaOH is introduced, the concentration of the acid is LESS than what you think it is

Answer 6

\[pH=4.74+\log \frac{ 0.1 }{ 0.4}=4.14\] what's my mistake?

Answer 7

you have to do the stoichiometry on the acid\base first, then use the LEFTOVER amounts in the KA equation

Answer 8

okay \[CH _{3}COOH+NaOH->CH _{3}COONa+H _{2}O\]
so, [HA]= 0.4-0.1=0.3?

Answer 9

okay thanks got it! :D

Answer 10

remember to use MOLES, not MOLARITY when you do stoichiometry like this, but yes