Question

When `"0.2 M"` of propionic acid dissolves in water, the `"pH"` was found to be `4.88`. Find the `K_a` of propionic acid?

Answer 1

This is an unrealistic answer, but I got `8.69 xx 10^(-10)`. The actual `K_a` is very different, at `1.34 xx 10^(-5)`!! The `"pH"` given was unrealistic, so the `K_a` calculated was unrealistic.

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By knowing the `"pH"`, you know the equilibrium concentration of `"H"^(+)`, which means you know what `x` is.

`"PropCOOH"(aq) + "H"_2"O"(l) rightleftharpoons "PropCOO"^(-)(aq) + "H"_3"O"^(+)(aq)`

`"I"" ""0.2 M"" "" "" "" "-" "" "" ""0 M"" "" "" "" "" ""0 M"`
`"C"" " - x" "" "" "" "-" "" "" "+x" "" "" "" "" "+x`
`"E"" "(0.2 - x) "M"" "-" "" "" "" "x" M"" "" "" "" "x" M"`

Since `"pH" = -log["H"_3"O"^(+)]`:

`10^(-"pH") = 10^(-4.88) = ["H"_3"O"^(+)] = x = 1.318 xx 10^(-5)` `"M"`

This gives you a `K_a` expression of:

`color(blue)(K_a) = (["PropCOO"^(-)]["H"_3"O"^(+)])/(["PropCOOH"])`

`= x^2/(0.2 - x)`

`= (1.318 xx 10^(-5))^2/(0.2 - 1.318 xx 10^(-5))`

`= color(blue)(8.69 xx 10^(-10))`

By the way, this `K_a`, although calculated correctly, is completely unrealistic. The actual `K_a` of propionic acid is around `1.34 xx 10^(-5)`... so the `"pH"` is unrealistic.

Let's calculate the `"pH"` using the correct `K_a` and compare. Since `K_a` is small (on the order of `10^(-5)` or less),

`K_a ~~ x^2/(0.2)`

`=> x ~~ sqrt(0.2K_a) = sqrt(0.2 * 1.34 xx 10^(-5)) = "0.001637 M"`

As a result, a more realistic `"pH"` would have been:

`color(red)("pH"_"realistic") = -log(0.001637) = color(red)(2.79)`,

which is quite a bit less than `4.88`...