Question

Question #b1714

Answer 1

`K_a=([H^+][A^-])/([HA])`

`ph =-log_10[H^+]`, so `[H^+]=10^(-pH)`

In this case, `[HA]=[CH_3CH_2COOH]=0.2 M` and `pH=2.79`

Because propanoic is a weak acid, most of the molecules won't dissociate, and so `[CH_3CH_2COO^-]` is roughly equal to `[CH_3CH_2COOH]`.

Now we can substitute `[H^+]` for `10^(-pH)` and `[CH_3CH_2COO^-]` for `[CH_3CH_2COOH]`, so we get the equation

`K_a=(10^(-pH)xx[CH_3CH_2COOH])/([CH_3CH_2COOH])=(10^(-2.79)xx0.20)/0.20=0.00162=1.62xx10^(-3)`