Question

Question #eae43

Answer 1

The percent ionization of salicylic acid is approximately `23.7%` according to the given data.

Explanation:

`(2g)/L * (mol)/(138g) approx 1.45*10^-2M`

`C_7H_6O_3(aq) rightleftharpoons H^(+)(aq)+C_7H_5O_3^(-)(aq)`

`K_a = ([H^+][C_7H_5O_3^(-)])/([C_7H_6O_3])`

`1.06*10^-3 = x^2/(0.0145 - x)`
`therefore x = [H^+] approx 0.00343M`

`%I = ([H^+])/([C_7H_6O_3]) = (0.00343)/(0.0145M)*100% approx 23.7%`

Answer 2

This is basically an equilibrium problem where you know the initial concentration (a usual one). But your cited `"pH"` is not accurate. I used your `K_a` to get a percent dissociation of `ul(17.7%)`.

[Your cited `"pH"` suggests a percent dissociation that is far too low.]

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Write the reaction first, so you have a way to organize:

`"HC"_7"H"_5"O"_3(aq) rightleftharpoons "C"_7"H"_5"O"_3^(-)(aq) + "H"^(+)(aq)`

`"I"" "["HC"_7"H"_5"O"_3]_i" "" "" ""0 M"" "" "" "" ""0 M"`
`"C"" "" "-x" "" "" "" "" "+x" "" "" "+x`
`"E"" "["HC"_7"H"_5"O"_3]_(i) - x" "x" M"" "" "" "" "x" M"`

METHOD 1: USING THE `bb("K"_a)` (this works)

Next, apparently we know `["HC"_7"H"_5"O"_3]_(i)` already (not equilibrium but initial). However, in water at `25^@ "C"`, the cited solubility is `"2.48 g/L"`, so that's what I'll use.

If you want to use a solubility, you must cite its temperature.

`color(green)(["HC"_7"H"_5"O"_3]_(i) = (2.48 cancel"g")/"L" xx "1 mol"/(138.12 cancel("g H"_7"H"_5"O"_3))`

`=` `color(green)("0.0179"_6 "M")`

Using your cited `K_a`, which is actually correct, we can do this in one of two ways. We know `["HC"_7"H"_5"O"_3]_(i)`, so...

`K_a = 1.06 xx 10^(-3) = x^2/("0.01796 M" - x)`

Solving for the quadratic form,

`x^2 + 1.06 xx 10^(-3) x - 1.06 xx 10^(-3) cdot 0.01796 = 0`,

and the result is `x = ul(["H"^(+)]_(eq) = "0.00387 M")` from the quadratic formula.

The percent dissociation is

`color(blue)(%"dissoc") = x/(["HA"]_i) xx 100%`

`= ("0.00387 M")/("0.0179"_6 "M" + "0.00387 M") xx 100%`

`= color(blue)ul(17.7%)`

And that's reasonable. Salicylic acid has a non-negligible percent dissociation with a `K_a` larger than `10^(-5)`.

METHOD 2: TRYING THE pH (this won't work)

With the `"pH"`,

`10^(-"pH") = ul(["H"^(+)]_(eq)) = x = 10^(-4.96) "M" = ul(1.10 xx 10^(-5) "M")`

As before, the solubility was `"0.01796 M"`.

This is pretty big in comparison to `x`. As a result, the percent dissociation using the cited `"pH"` will be far too low.

`color(red)(%"dissoc") = x/(["HA"]_i) xx 100%`

`= (10^(-4.96) "M")/("0.0179"_6 "M") xx 100% = color(red)(0.0611%)`

which doesn't make sense how low this is, given the size of `K_a`.

In fact, the `K_a` obtained from this percent dissociation is

`color(red)(K_a) = (10^(-4.96) "M")^2/("0.0179"_6 "M" - 10^(-4.96)"M") = color(red)(6.70 xx 10^(-9))`,

which is far from the actual.