If it helps, I have:
[H+] = 0.00115
[OH-] = 8.696e-12
pH = 2.94
These are verified to be correct.
For propanoic acid HC3H5O2, Ka = 1.3 ✕ 10−5 and 0.102 M solution.
[HC3H5O2] ____M
[C3H5O2− ] ____M
percent dissociation ____percent
3 answers
So you have the answers and they are verified. What do you need from us?
[H⁺] = [C₃H₅O₂ˉ] = √Ka[HC₃H₅O₂]) = √(1.3 x 10ˉ⁵)(0.102) M = 0.00115M
[HC₃H₅O₂] = 0.102M
%Dissociation = ([H⁺] or [C₃H₅O₂ˉ])]/[HC₃H₅O₂])100% = [(0.00115M)/(0.102M)]100% = 1.13% dissociated
[HC₃H₅O₂] = 0.102M
%Dissociation = ([H⁺] or [C₃H₅O₂ˉ])]/[HC₃H₅O₂])100% = [(0.00115M)/(0.102M)]100% = 1.13% dissociated
1 answer