# Question #eae43

##### 2 Answers

The percent ionization of salicylic acid is approximately

#### Explanation:

This is basically an equilibrium problem where you know the initial concentration (a usual one). But your cited

[Your cited

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

*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 **actually correct**, we can do this in one of two ways. We know

#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

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

With the

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

As before, the solubility was

This is pretty big in comparison to **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

#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

farfrom the actual.