[noname123]

60 mL 0.35 M acetic acid solution and 50 mL 1.10 M sodium hydroxide solution are run from burettes into a 200.00 mL volumetric flask, and the solution made up to the mark. What is the pH of the solution? (Ka for acetic acid 1.75 x 10-5)


well this is how i approached this problem..

first find which one is limiting reagent...

then we can find the [H+] concentration using Ka = [H+] [CH₃COO^-] devide by CH₃COOH

then do i find the dilution factor ??..

leme show it its easier..

     OH⁺  +  HC₂H₃O₂  = C₂H₃O^-  + H₂O


before reaction mole OH = 50mL X 1.10
                                = 55m mol
                                = 0.055mol

before reaction mole HC₂H₃O₂ = 60 X 0.35
                                                                                       = 0.021mol

therefore HC₂H₃O₂ is limiting

then [OH-] = 0.034 [ i got this from 0.055-0.021]
                      (0.06+0.05+0.09) [total volume]


then use K_w = K_a X K_b to find [H⁺]

am i doing the right thing??

[AWK]

[OH-]=34/200=0.17
You have an excess of strong base, and you can ignore other equilibriums except Kw
Then
Kw=[H+][OH-]

[noname123]

then that means im not using the given K_a for acetic acid..

is that just a trick in the question..?

[AWK]

In the excess of strong base (or acid) you can ignore all weak species.